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Class 12 Maths Chapter 7 Integrals(INDEFINITE INTEGRALS) EXERCISE 7.1 TO EXERCISE 7.7

Exercise 7.1

Find an antiderivative (or integral) of the following by the method of inspection:

Question 1.
sin 2x
Solution:
\int { sin2x\quad dx=-\frac { cos2x }{ 2 } +C }

Question 2.
cos 3x
Solution:
\int { cos3x\quad dx=\frac { sin3x }{ 3 } +C }

Question 3.
{ e }^{ 2x }
Solution:
\int { { e }^{ 2x }dx=\frac { { e }^{ 2x } }{ 2 } +C }

Question 4.
(ax + c)²
Solution:
\int { { (ax+b) }^{ 2 }dx=\frac { { (ax+b) }^{ 3 } }{ 3a } } +C

Question 5.
{ sin\quad 2x-4e }^{ 3x }
Solution:
\int { \left( { sin2x-4e }^{ 3x } \right) dx=-\frac { cos2x }{ 2 } -\frac { { 4e }^{ 3x } }{ 3 } +C }

Find the following integrals in Exercises 6 to 20 :

Question 6.
\int { \left( { 4e }^{ 3x }+1 \right) dx }
Solution:
=\int { { 4e }^{ 3x }dx+\int { dx=\frac { 4 }{ 3 } { e }^{ 3x }+x+c } }

Question 7.
\int { { x }^{ 2 }\left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx }
Solution:
=\int { { x }^{ 2 }\left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx } =\frac { { x }^{ 3 } }{ 3 } -x+C

Question 8.
\int { { (ax }^{ 2 }+bx+c)dx }
Solution:
=\frac { { ax }^{ 3 } }{ 3 } +\frac { { bx }^{ 2 } }{ 2 } +cx+d

Question 9.
\int { \left( { 2x }^{ 2 }+{ e }^{ x } \right) dx }
Solution:
=\frac { { 2x }^{ 3 } }{ 3 } +{ e }^{ x }+c

Question 10.
\int { { \left[ \sqrt { x } -\frac { 1 }{ \sqrt { x } } \right] }^{ 2 }dx }
Solution:
=\frac { { x }^{ 2 } }{ 2 } +logx-2x+C

Question 11.
\int { \frac { { x }^{ 3 }+{ 5x }^{ 2 }-4 }{ { x }^{ 2 } } dx }
Solution:
\int { \left( \frac { { x }^{ 3 } }{ { x }^{ 2 } } +\frac { { 5x }^{ 2 } }{ { x }^{ 2 } } -\frac { 4 }{ { x }^{ 2 } } \right) }
=\int { xdx+5\int { 1dx-4 } \int { { x }^{ 2 }dx } }
=\frac { { x }^{ 2 } }{ 2 } +5x+\frac { 4 }{ x } +c

Question 12.
\int { \frac { { x }^{ 3 }+3x+4 }{ \sqrt { x } } dx }
Solution:
=\int { \left( { x }^{ \frac { 5 }{ 2 } }+{ 3x }^{ \frac { 1 }{ 2 } }+4{ x }^{ -\frac { 1 }{ 2 } } \right) } dx
=\frac { 2 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+{ 2x }^{ \frac { 3 }{ 2 } }+8\sqrt { x } +c

Question 13.
\int { \frac { { x }^{ 3 }-{ x }^{ 2 }+x-1 }{ x-1 } dx }
Solution:
=\int { \frac { { x }^{ 2 }(x-1)+(x-1) }{ x-1 } dx }
=\int { \left( { x }^{ 2 }+1 \right) dx } =\frac { { x }^{ 3 } }{ 3 } +x+c

Question 14.
\int { \left( 1-x \right) \sqrt { x } dx }
Solution:
=\int { { x }^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 3 }{ 2 } }dx\quad =\quad \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { 2 }{ 5 } { x }^{ \frac { 5 }{ 2 } } }

Question 15.
\int { \sqrt { x } \left( { 3x }^{ 2 }+2x+3 \right) dx }
Solution:
=\int { \left( { 3x }^{ \frac { 5 }{ 2 } }+{ 2 }^{ \frac { 3 }{ 2 } }+{ 3x }^{ \frac { 1 }{ 2 } } \right) dx }
=\frac { 6 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+\frac { 4 }{ 5 } { x }^{ \frac { 5 }{ 2 } }+\frac { 6 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+c

Question 16.
\int { (2x-3cosx+{ e }^{ x })dx }
Solution:
=\frac { { 2x }^{ 2 } }{ 2 } -3sinx+{ e }^{ x }+c
={ x }^{ 2 }-3sinx+{ e }^{ x }+c

Question 17.
\int { \left( { 2x }^{ 2 }-3sinx+5\sqrt { x } \right) dx }
Solution:
=\frac { { 2x }^{ 3 } }{ 3 } +3cosx+5\frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } +c
=\frac { 2 }{ 3 } { x }^{ 3 }+3cosx+\frac { 10 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+c

Question 18.
\int { secx(secx+tanx)dx }
Solution:
=\int { { (sec }^{ 2 }x+secxtanx)dx }
= tanx + secx + c

Question 19.
\int { \frac { { sec }^{ 2 }x }{ { cosec }^{ 2 }x } dx }
Solution:
=\int { \frac { 1 }{ { cos }^{ 2 }x } } { sin }^{ 2 }xdx
=\int { tan } ^{ 2 }xdx\quad =\int { { (sec }^{ 2 }x-1)dx\quad =tanx-x+c }

Question 20.
\int { \frac { 2-3sinx }{ { cos }^{ 2 }x } dx }
Solution:
=\int { \left( \frac { 2 }{ { cos }^{ 2 }x } -3\frac { sinx }{ { cos }^{ 2 }x } \right) dx }
=\int { ({ 2sec }^{ 2 }x-3secxtanx)dx }
= 2tanx – 3secx + c

Choose the correct answer in Exercises 21 and 22.

Question 21.
The antiderivative \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) equals
(a) \frac { 1 }{ 3 } { x }^{ \frac { 1 }{ 3 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c
(b) \frac { 2 }{ 3 } { x }^{ \frac { 2 }{ 3 } }+{ \frac { 1 }{ 2 } x }^{ 2 }+c
(c) \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c
(d) \frac { 3 }{ 2 } { x }^{ \frac { 3 }{ 2 } }+\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } }+c
Solution:
(c) \int { \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) dx }
=\int { \left( { x }^{ \frac { 1 }{ 2 } }+{ x }^{ \frac { 1 }{ 2 } } \right) dx }
=\frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c

Question 22.
If \frac { d }{ dx } f(x)={ 4x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } } such that f(2)=0 then f(x) is
(a) { x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } -\frac { 129 }{ 8 }
(b) { x }^{ 3 }+\frac { 1 }{ { x }^{ 4 } } +\frac { 129 }{ 8 }
(c) { x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } +\frac { 129 }{ 8 }
(d) { x }^{ 3 }+\frac { 1 }{ { x }^{ 4 } } -\frac { 129 }{ 8 }
Solution:
(a) f(x)=\int { \left( { 4x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } } \right) dx }
={ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } +c
\therefore f(2)={ (2) }^{ 4 }+\frac { 1 }{ { (2) }^{ 3 } } +c=0=-\frac { 129 }{ 8 }

Exercise 7.2

Integrate the functions in Exercises 1 to 37:

Question 1.
\frac { 2x }{ 1+{ x }^{ 2 } }
Solution:
Let 1+x² = t
⇒ 2xdx = dt
\therefore \int { \frac { 2x }{ 1+{ x }^{ 2 } } dx\quad = } \int { \frac { dt }{ t } \quad =logt+C\quad =log(1+{ x }^{ 2 })+C }

Question 2.
\frac { { \left( logx \right) }^{ 2 } }{ x }
Solution:
Let logx = t
⇒ \frac { 1 }{ x }dx=dt
\therefore \int { \frac { { (logx) }^{ 2 } }{ x } dx } \quad =\int { { t }^{ 2 }dt } \quad =\frac { { t }^{ 3 } }{ 3 } +c\quad =\frac { 1 }{ 3 } { (logx) }^{ 3 }+c

Question 3.
\frac { 1 }{ x+xlogx }
Solution:
Put 1+logx = t
∴ \frac { 1 }{ x }dx=dt
\int { \frac { 1 }{ x(1+logx) } dx } =\int { \frac { 1 }{ t } dt } =log|t|+c
= log|1+logx|+c

Question 4.
sinx sin(cosx)
Solution:
Put cosx = t, -sinx dx = dt
\int { sinx\quad sin(cosx)dx } =-\int { sin(cosx) } (-sinx)dx
=-\int { sint\quad dt } \quad =cost+c\quad =cos(cosx)+c

Question 5.
sin(ax+b) cos(ax+b)
Solution:
let sin(ax+b) = t
⇒ cos(ax+b)dx = dt
\therefore \int { sin(ax+b)cos(ax+b)dx } =\frac { 1 }{ a } \int { t\quad dt }
=\frac { 1 }{ a } .\frac { { t }^{ 2 } }{ 2 } +c\quad =\frac { 1 }{ 2a } { sin }^{ 2 }(ax+b)+C

Question 6.
\sqrt { ax+b }
Solution:
\int { \sqrt { ax+b } dx } \quad =\frac { 2 }{ 3a } { (ax+b) }^{ \frac { 3 }{ 2 } }+C

Question 7.
x\sqrt { x+2 }
Solution:
Let x+2 = t²
⇒ dx = 2t dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 7

Question 8.
x\sqrt { 1+{ 2x }^{ 2 } }
Solution:
let 1+2x² = t²
⇒ 4x dx = 2t dt
x\quad dx=\frac { t }{ 2 } dt\int { x\sqrt { 1+{ 2x }^{ 2 } } dx }
=\frac { 1 }{ 2 } \int { { t }^{ 2 }dt } =\frac { { t }^{ 3 } }{ 6 } +c=\frac { 1 }{ 6 } { ({ 1+2x }^{ 2 }) }^{ \frac { 3 }{ 2 } }+c

Question 9.
(4x+2)\sqrt { { x }^{ 2 }+x+1 }
Solution:
let x²+x+1 = t
⇒(2x+1)dx = dt
\therefore \int { (4x+1)\sqrt { { x }^{ 2 }+x+1 } dx } =2\int { \sqrt { t } dt }
=\frac { { 2t }^{ \frac { 3 }{ 2 } } }{ ^{ \frac { 3 }{ 2 } } } +c\quad =\frac { 4 }{ 3 } { t }^{ \frac { 3 }{ 2 } }+c\quad =\frac { 4 }{ 3 } { ({ x }^{ 2 }+x+1) }^{ \frac { 3 }{ 2 } }+c

Question 10.
\frac { 1 }{ x-\sqrt { x } }
Solution:
\int { \frac { 1 }{ x-\sqrt { x } } dx } =\int { \frac { 1 }{ \sqrt { x } (\sqrt { x-1 } ) } dx } =I
Let √x-1 = t
\frac { 1 }{ 2 } { x }^{ -\frac { 1 }{ 2 } }dx=dt
I=2\int { \frac { dt }{ t } }
= 2logt + c
= 2log(√x-1)+c

Question 11.
\frac { x }{ \sqrt { x+4 } } ,x>0
Solution:
let x+4 = t
⇒ dx = dt, x = t-4
byjus class 12 maths Chapter 7 Integrals 11

Question 12.
{ { (x }^{ 3 }-1) }^{ \frac { 1 }{ 3 } }.{ x }^{ 5 }
Solution:
\int { { { (x }^{ 3 }-1) }^{ \frac { 1 }{ 3 } }.{ x }^{ 5 }.dx } \quad =\frac { 1 }{ 7 } { { (x }^{ 3 }-1) }^{ \frac { 7 }{ 3 } }+\frac { 1 }{ 4 } { { (x }^{ 3 }-1) }^{ \frac { 4 }{ 3 } }+c

Question 13.
\frac { { x }^{ 2 } }{ { { (2+3x }^{ 3 }) }^{ 3 } }
Solution:
Let 2+3x³ = t
⇒ 9x² dx = dt
\therefore \int { \frac { { x }^{ 2 } }{ { { (2+3x }^{ 3 }) }^{ 3 } } dx } =\frac { 1 }{ 9 } \int { \frac { dt }{ { t }^{ 3 } } =\frac { 1 }{ 9 } \int { { t }^{ -3 }dt } }
=-\frac { 1 }{ { 18t }^{ 2 } } +c\quad =-\frac { 1 }{ 18(2+{ 3x }^{ 3 })^{ 2 } } +c

Question 14.
\frac { 1 }{ x(logx)^{ m } } ,x>0
Solution:
Put log x = t, so that \frac { 1 }{ x }dx=dt
\therefore \int { \frac { 1 }{ { x(logx) }^{ m } } dx } =\int { \frac { dt }{ { t }^{ m } } =\frac { { t }^{ -m+1 } }{ -m+1 } +c }
=\frac { { (logx) }^{ 1-m } }{ 1-m } +c

Question 15.
\frac { x }{ 9-4{ x }^{ 2 } }
Solution:
put 9-4x² = t, so that -8x dx = dt
\therefore \int { \frac { x }{ 9-{ 4x }^{ 2 } } dx } =-\frac { 1 }{ 8 } \int { \frac { dt }{ t } } =-\frac { 1 }{ 8 } log|t|+c
=\frac { 1 }{ 8 } log\frac { 1 }{ |9-{ 4x }^{ 2 }| } +c

Question 16.
{ e }^{ 2x+3 }
Solution:
put 2x+3 = t
so that 2dx = dt
\int { { e }^{ 2x+3 } } dx\quad =\frac { 1 }{ 2 } \int { { e }^{ t }dt } \quad =\frac { 1 }{ 2 } { e }^{ t }+c\quad =\frac { 1 }{ 2 } { e }^{ 2x+3 }+c

Question 17.
\frac { x }{ { e }^{ { x }^{ 2 } } }
Solution:
Let x² = t
⇒ 2xdx = dt ⇒ xdx=\frac { dt }{ 2 }
\therefore \int { \frac { x }{ { e }^{ { x }^{ 2 } } } dx } \quad =\frac { 1 }{ 2 } \int { \frac { dt }{ { e }^{ t } } \quad =\frac { 1 }{ 2 } \int { { e }^{ -t } } dt }
=-\frac { 1 }{ 2 } { e }^{ { -x }^{ 2 } }+c

Question 18.
\frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } }
Solution:
let\quad { tan }^{ -1 }x=t\Rightarrow \frac { 1 }{ 1+{ x }^{ 2 } } dx=dt
\therefore \int { \frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } } dx } \quad =\int { { e }^{ t }dt\quad ={ e }^{ { tan }^{ -1 }x }+c }

Question 19.
\frac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 }
Solution:
\int { \frac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 } dx\quad =\int { \frac { { e }^{ x }({ e }^{ x }-{ e }^{ -x }) }{ { e }^{ x }({ e }^{ x }+{ e }^{ -x }) } dx=I } }
put ex+e-x = t
so that (ex-e-x)dx = dt
\therefore I=\int { \frac { dt }{ t } =log|t|+c } =log|{ e }^{ x }+{ e }^{ -x }|+c

Question 20.
\frac { { e }^{ 2x }-{ e }^{ 2x } }{ { e }^{ 2x }+{ e }^{ -2x } }
Solution:
put e2x-e-2x = t
so that (2e2x-2e-2x)dx = dt
\therefore \int { \frac { { e }^{ 2x }-{ e }^{ 2x } }{ { e }^{ 2x }+{ e }^{ -2x } } } dx=\frac { 1 }{ 2 } \int { \frac { 1 }{ t } dt } =\frac { 1 }{ 2 } log|t|+c
=\frac { 1 }{ 2 } log+|{ e }^{ 2x }+{ e }^{ -2x }|+c

Question 21.
tan²(2x-3)
Solution:
∫tan²(2x-3)dx = ∫[sec²(2x-3)-1]dx = I
put 2x-3 = t
so that 2dx = dt
I = \frac { 1 }{ 2 } ∫sec²t dt-x+c
= \frac { 1 }{ 2 }t-x+c
= \frac { 1 }{ 2 }tan(2x-3)-x+c

Question 22.
sec²(7-4x)
Solution:
∫sec²(7-4x)dx
= \frac { tan(7-4x) }{ -4 }+c

Question 23.
\frac { { sin }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } }
Solution:
let\quad { sin }^{ -1 }x=t\quad \Rightarrow \frac { 1dx }{ \sqrt { 1-{ x }^{ 2 } } } =dt
\int { \frac { { sin }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } dx } =\int { t\quad dt } =\frac { 1 }{ 2 } { t }^{ 2 }+c=\frac { 1 }{ 2 } { { (sin }^{ -1 }x) }^{ 2 }+c

Question 24.
\frac { 2cosx-3sinx }{ 6cosx+4sinx }
Solution:
put 2sinx+4cosx = t
⇒ (2cosx-3sinx)dx = dt
\frac { 1 }{ 2 } \int { \frac { 2cosx-3sinx }{ 2sinx+3cosx } dx } =\frac { 1 }{ 2 } \int { \frac { dt }{ t } } =\frac { 1 }{ 2 } log|t|+c
\frac { 1 }{ 2 } log|2sinx+3cosx|+c

Question 25.
\frac { 1 }{ { cos }^{ 2 }x{ (1-tanx) }^{ 2 } }
Solution:
put 1-tanx = t
so that -sec²x dx = dt
\therefore \int { \frac { 1 }{ { cos }^{ 2 }x{ (1-tanx) }^{ 2 } } dx } =\int { \frac { { sec }^{ 2 }x }{ { (1-tanx) }^{ 2 } } dx }
=-\int { \frac { dt }{ { t }^{ 2 } } } =\frac { 1 }{ t } +c=\frac { 1 }{ (1-tanx) } +c

Question 26.
\frac { cos\sqrt { x } }{ \sqrt { x } }
Solution:
put\quad \sqrt { x } =t,so\quad that\frac { 1 }{ 2\sqrt { x } } dx=dt
\therefore \int { \frac { cos\sqrt { x } }{ \sqrt { x } } } dx\quad =\quad 2\quad =\int { cost\quad dt\quad = } 2sint+c
= 2sin√x+c

Question 27.
\sqrt { sin2x } cos2x
Solution:
put sin2x = t²
⇒ cos2x dx = t dt
\therefore \int { \sqrt { sin2x } .cos2x\quad dx } \quad =\int { t.tdt=\frac { { t }^{ 3 } }{ 3 } +c }
=\frac { { (sin2x) }^{ \frac { 3 }{ 2 } } }{ 3 } +c

Question 28.
\frac { cosx }{ \sqrt { 1+sinx } }
Solution:
put 1+sinx = t²
⇒cosx dx = 2t dt
\therefore \int { \frac { cosx }{ \sqrt { 1+sinx } } dx } =2\int { dt } =2t+c
=2\sqrt { 1+sinx } +c

Question 29.
cotx log sinx
Solution:
put log sinx = t,
⇒ cot x dx = dt
\therefore \int { cot\quad logsinx\quad dx } =\int { t } dt\quad =\frac { { t }^{ 2 } }{ 2 } +c
=\frac { 1 }{ 2 } { (log\quad sinx) }^{ 2 }+c

Question 30.
\frac { sinx }{ 1+cosx }
Solution:
put 1+cosx = t
⇒ -sinx dx = dt
\therefore \int { \frac { sinx }{ 1+cosx } dx } =\int { -\frac { dt }{ t } } =-logt+c
=-log(1+cosx)+c

Question 31.
\frac { sinx }{ { (1+cosx) }^{ 2 } }
Solution:
put 1+cosx = t
so that -sinx dx = dt
\therefore \int { \frac { sinx }{ { (1+cosx) }^{ 2 } } dx } =-\int { \frac { dt }{ { t }^{ 2 } } }
=\frac { 1 }{ t } +c=\frac { 1 }{ 1+cosx } +c

Question 32.
\frac { 1 }{ 1+cotx }
Solution:
\int { \frac { 1 }{ 1+\frac { cosx }{ sinx } } } dx=\frac { 1 }{ 2 } \int { \frac { 2sinx\quad dx }{ sinx+cosx } }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 32

Question 33.
\frac { 1 }{ 1-tanx }
Solution:
\int { \frac { 1 }{ 1-tanx } } dx=\frac { 1 }{ 2 } \int { \frac { 2cosx\quad dx }{ cosx-sinx } }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 33

Question 34.
\frac { \sqrt { tanx } }{ sinxcosx }
Solution:
\int { \frac { \sqrt { tanx } }{ sinxcosx } dx } =\int { \frac { \sqrt { tanx } }{ tanx } } .{ sec }^{ 2 }xdx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 34

Question 35.
\frac { { (1+logx) }^{ 2 } }{ x }
Solution:
let 1+logx = t
⇒ \frac { 1 }{ x }dx=dt
\int { \frac { { (1+logx) }^{ 2 } }{ x } dx } =\int { { t }^{ 2 }dt } =\frac { { t }^{ 3 } }{ 3 } +c
=\frac { 1 }{ 3 } { (1+logx) }^{ 3 }+c

Question 36.
\frac { (x+1){ (x+logx) }^{ 2 } }{ x }
Solution:
put x+logx = t
\left( \frac { x+1 }{ x } \right) dx=dt
\therefore \int { \frac { { (x+1)(x+logx) }^{ 2 } }{ x } } dx=\int { { t }^{ 2 }dt }
=\frac { { (x+logx) }^{ 3 } }{ 3 } +c

Question 37.
\frac { { x }^{ 3 }sin({ tan }^{ -1 }{ x }^{ 4 }) }{ 1+{ x }^{ 8 } } dx
Solution:
put\quad { tan }^{ -1 }{ x }^{ 4 }=t\quad so\quad that\frac { 1 }{ 1+{ x }^{ 8 } } .{ 4x }^{ 3 }dx=dt
\therefore \int { \frac { { x }^{ 3 }sin({ tan }^{ -1 }{ x }^{ 4 }) }{ 1+{ x }^{ 8 } } } dx=\frac { 1 }{ 4 } \int { sint\quad dt }
=\frac { 1 }{ 4 } (-cost)+c=-\frac { 1 }{ 4 } cos({ tan }^{ -1 }{ x }^{ 4 })+c

Choose the correct answer in exercises 38 and 39

Question 38.
\int { \frac { { 10x }^{ 9 }+{ 10 }^{ x }log{ e }^{ 10 } }{ { x }^{ 10 }+{ 10 }^{ x } } dx }
(a) 10x – x10 + C
(b) 10x + x10 + C
(c) (10x – x10) + C
(d) log (10x + x10) + C
Solution:
(d) \int { \frac { { 10x }^{ 9 }+{ 10 }^{ x }log{ e }^{ 10 } }{ { x }^{ 10 }+{ 10 }^{ x } } dx }
= log (10x + x10) + C

Question 39.
\int { \frac { dx }{ { sin }^{ 2 }x{ \quad cos }^{ 2 }x } = }
(a) tanx + cotx + c
(b) tanx – cotx + c
(c) tanx cotx + c
(d) tanx – cot2x + c
Solution:
(c) \int { \frac { dx }{ { sin }^{ 2 }x{ \quad cos }^{ 2 }x } = }
=\int { \left( { sec }^{ 2 }x+{ cosec }^{ 2 }x \right) dx }
= tanx – cotx + c

We hope the NCERT Solutions for Class 12 Maths

Exercise 7.3

Find the integrals of the functions in Exercises 1 to 22.

Question 1.
sin²(2x+5)
Solution:
∫sin²(2x+5)dx
= \frac { 1 }{ 2 }∫[1-cos2(2x+5)]dx
= \frac { 1 }{ 2 }∫[1-cos(4x+10)]dx
= \frac { 1 }{ 2 } \left[ x-\frac { sin(4x+10) }{ 4 } \right] +c

Question 2.
sin3x cos4x
Solution:
∫sin3x cos4x
= \frac { 1 }{ 2 }∫[sin(3x+4x)+cos(3x-4x)]dx
= \frac { 1 }{ 2 }∫[sin7x+sin(-x)]dx
= -\frac { 1 }{ 14 } cos7x+\frac { 1 }{ 2 } cosx+c

Question 3.
∫cos2x cos4x cos6x dx
Solution:
\frac { 1 }{ 2 } ∫cos2x cos4x cos6x dx
= \frac { 1 }{ 2 } ∫(cos6x+cos2x) cos6x dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 3

Question 4.
∫sin3(2x+1)dx
Solution:
= \frac { 1 }{ 4 } ∫[3sin(2x+1)-sin3(2x+1)]dx
= -\frac { 3 }{ 8 } cos(2x+1)+\frac { 1 }{ 24 } [4{ cos }^{ 3 }(2x+1)-3cos(2x+1)]+c
= -\frac { 1 }{ 2 } cos(2x+1)+\frac { 1 }{ 6 } { cos }^{ 3 }(2x+1)+c

Question 5.
sin3x cos3x
Solution:
put sin x = t
⇒ cos x dx = dt
\therefore \int { { sin }^{ 3 }x{ cos }^{ 3 }xdx } =\int { { t }^{ 3 }(1-{ t }^{ 2 })dt }
\frac { { t }^{ 4 } }{ 4 } -\frac { { t }^{ 6 } }{ 6 } +c=\frac { { (sinx) }^{ 4 } }{ 4 } -\frac { { (sinx) }^{ 6 } }{ 6 } +c

Question 6.
sinx sin2x sin3x
Solution:
∫sinx sin2x sin3x dx
= \frac { 1 }{ 2 } ∫ 2sin x sin 2x sin 3x dx
= \frac { 1 }{ 2 } ∫ (cosx – cos3x)sin 3x dx
= \frac { 1 }{ 2 } ∫ (sin 4x + sin 2x – sin 6x)dx
= \frac { 1 }{ 4 } \left\{ \frac { -cos4x }{ 4 } -\frac { cos2x }{ 2 } +\frac { cos6x }{ 6 } \right\} +c

Question 7.
sin 4x sin 8x
Solution:
\frac { 1 }{ 2 }∫sin 4x sin 8xdx
= \frac { 1 }{ 2 }∫(cos 4x – cos 12x)dx
= \frac { 1 }{ 2 } \left[ \frac { sin4x }{ 4 } -\frac { sin12x }{ 12 } \right] +c

Question 8.
\frac { 1-cosx }{ 1+cosx }
Solution:
\int { \frac { 1-cosx }{ 1+cosx } dx }
\int { \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { 2cos }^{ 2 }\frac { x }{ 2 } } dx } =\int { { tan }^{ 2 }\frac { x }{ 2 } dx }
=\int { \left[ { sec }^{ 2 }\frac { x }{ 2 } -1 \right] } dx\quad =2tan\frac { x }{ 2 } -x+c

Question 9.
\frac { cosx }{ 1+cosx }
Solution:
\int { \frac { cosx }{ 1+cosx } dx }
=\int { 1 } dx-\int { \frac { 1 }{ 1+cosx } dx }
=x-\frac { 1 }{ 2 } \int { { sec }^{ 2 }\frac { x }{ 2 } dx+c\quad =x-tan\frac { x }{ 2 } +c }

Question 10.
∫sinx4 dx
Solution:
\int { { (\frac { 1-cos2x }{ 2 } ) }^{ 2 }dx } \quad =\frac { 1 }{ 4 } \int { \left( { 1+cos }^{ 2 }2x-2cos2x \right) dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 10

Question 11.
cos4 2x
Solution:
∫ cos4 2x dx
\int { { \left( \frac { 1+cos4x }{ 2 } \right) }^{ 2 } } dx
tiwari academy class 12 maths Chapter 7 Integrals 11

Question 12.
\frac { { sin }^{ 2 }x }{ 1+cosx }
Solution:
\int { \frac { { sin }^{ 2 }x }{ 1+cosx } } dx\quad =\int { \frac { 1-{ cos }^{ 2 }x }{ 1+cosx } } dx
\int { (1-cosx) } dx\quad =x-sinx+c

Question 13.
\frac { cos2x-cos2\alpha }{ cosx-cos\alpha }
Solution:
let I = \int { \frac { \left( { 2cos }^{ 2 }x-1 \right) -\left( { 2cos }^{ 2 }\alpha -1 \right) }{ cosx-cos\alpha } } dx
\int { \frac { 2\left( { cos }x-cos\alpha \right) -\left( { cos }x+cos\alpha \right) }{ cosx-cos\alpha } } dx
= 2∫cos x dx + 2cos α∫dx
= 2(sinx+xcosα)+c

Question 14.
\frac { cosx-sinx }{ 1+sin2x }
Solution:
let I = \int { \frac { cosx-sinx }{ 1+sin2x } } dx=\int { \frac { cosx-sinx }{ { (cosx+sinx) }^{ 2 } } dx }
put cosx+sinx = t
⇒ (-sinx+cosx)dx = dt
I=\int { \frac { dt }{ { t }^{ 2 } } } =-\frac { 1 }{ t } +c\quad =\frac { -1 }{ cosx+sinx } +c

Question 15.
\int { { tan }^{ 3 }2x\quad sec2x\quad dx=I }
Solution:
I = ∫(sec22x-1)sec2x tan 2xdx
put sec2x=t,2 sec2x tan2x dx=dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 15

Question 16.
tan4x
Solution:
let I = ∫tan4 dx
= ∫(sec²x-1)²dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 16

Question 17.
\frac { { sin }^{ 3 }x+{ cos }^{ 3 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x }
Solution:
\int { \left( \frac { { sin }^{ 3 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } +\frac { { cos }^{ 2 }x }{ sinx{ cos }^{ 2 }x } \right) dx }
= secx-cosecx+c

Question 18.
\frac { cos2x+{ 2sin }^{ 2 }x }{ { cos }^{ 2 }x }
Solution:
I=\int { \frac { \left( { cos }^{ 2 }x-{ sin }^{ 2 }x \right) +2{ sin }^{ 2 }x }{ { cos }^{ 2 }x } } dx
=\int { \frac { \left( { cos }^{ 2 }x-{ sin }^{ 2 }x \right) }{ { cos }^{ 2 }x } } dx\quad =\int { { sec }^{ 2 }xdx\quad =tanx+c }

Question 19.
\frac { 1 }{ sinx{ cos }^{ 3 }x }
Solution:
I=\int { \left( tanx+\frac { 1 }{ tanx } \right) } { sec }^{ 2 }xdx
put tanx = t
so that sec²x dx = dt
I=\int { \left( t+\frac { 1 }{ t } \right) } dt\quad =\frac { { t }^{ 2 } }{ 2 } +log|t|+c
=log|tanx|+\frac { 1 }{ 2 } { tan }^{ 2 }x+c

Question 20.
\frac { cos2x }{ { (cosx+sinx) }^{ 2 } }
Solution:
I=\int { \frac { { cos }^{ 2 }x-{ sin }^{ 2 }x }{ (cosx+sinx)^{ 2 } } dx } =\int { \frac { cosx-sinx }{ cosx+sinx } dx }
put cosx+sinx=t
⇒(-sinx+cox)dx = dt
I=\int { \frac { dt }{ t } } =log|t|+c\quad =log|cosx+sinx|+c

Question 21.
sin-1 (cos x)
Solution:
\int { { sin }^{ -1 }(cosx)dx } \quad ={ sin }^{ -1 }\left[ sin\left( \frac { \pi }{ 2 } -x \right) \right] dx
\int { \left( \frac { \pi }{ 2 } -x \right) dx } \quad =\frac { \pi x }{ 2 } -\frac { { x }^{ 2 } }{ 2 } +c

Question 22.
\int { \frac { 1 }{ cos(x-a)cos(x-b) } dx }
Solution:
\frac { 1 }{ sin(a-b) } \int { \frac { sin[(x-b)-(x-a)] }{ cos(x-a)cos(x-b) } dx }
=\frac { 1 }{ sin(a-b) } \left[ \int { tan(x-b)dx-\int { tan(x-a)dx } } \right]
=\frac { 1 }{ sin(a-b) } log\left| \frac { cos(x-a) }{ cos(x-b) } \right| +c

Question 23.
\int { \frac { { sin }^{ 2 }x-{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx\quad is\quad equal\quad to
(a) tanx+cotx+c
(b) tanx+cosecx+c
(c) -tanx+cotx+c
(d) tanx+secx+c
Solution:
(a) \int { \frac { { sin }^{ 2 }x-{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx
= ∫(sec²x-cosec²x)dx
= tanx+cotx+c

Question 24.
\int { \frac { e^{ x }(1+x) }{ cos^{ 2 }({ e }^{ x }.{ x }) } } dx\quad is\quad equal\quad to
(a) -cot(e.xx)+c
(b) tan(xex)+c
(c) tan(ex)+c
(d) cot ex+c
Solution:
(b) \int { \frac { e^{ x }(1+x) }{ cos^{ 2 }({ e }^{ x }.{ x }) } } dx
= ∫sec²t dt
= tan t+c = tan(xex)+c

Exercise 7.4

Integrate the functions in exercises 1 to 23

Question 1.
\frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 }
Solution:
Let x3 = t ⇒ 3x²dx = dt
\int { \frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 } dx } =\int { \frac { dt }{ { t }^{ 2 }+1 } } ={ tan }^{ -1 }t+c
= tan-1 (x3)+c

Question 2.
\frac { 1 }{ \sqrt { 1+{ 4x }^{ 2 } } }
Solution:
\frac { 1 }{ 2 } \int { \frac { dx }{ \sqrt { \frac { 1 }{ 4 } +{ x }^{ 2 } } } } =\frac { 1 }{ 2 } \int { \frac { dx }{ \sqrt { { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ x }^{ 2 } } } }
=\frac { 1 }{ 2 } log\left| 2x+\sqrt { 1+{ 4x }^{ 2 } } \right| +c

Question 3.
\frac { 1 }{ \sqrt { { (2-x) }^{ 2 }+1 } }
Solution:
put (2-x)=t
so that -dx=dt
⇒ dx=-dt
\int { \frac { dx }{ \sqrt { { (2-x) }^{ 2 }+1 } } } =-\int { \frac { dt }{ \sqrt { { t }^{ 2 }+1 } } } =-log|t+\sqrt { { t }^{ 2 }+1 } |+c
=log\left| \frac { 1 }{ (2-x)+\sqrt { { x }^{ 2 }-4x+5 } } \right| +c

Question 4.
\frac { 1 }{ \sqrt { 9-{ 25x }^{ 2 } } }
Solution:
\int { \frac { dx }{ \sqrt { 9-{ 25x }^{ 2 } } } } =\frac { 1 }{ 5 } \int { \frac { dx }{ \sqrt { { \left( \frac { 3 }{ 5 } \right) }^{ 2 }-{ x }^{ 2 } } } }
=\frac { 1 }{ 5 } { sin }^{ -1 }\left( \frac { x }{ \frac { 3 }{ 5 } } \right) +c\quad =\frac { 1 }{ 5 } { sin }^{ -1 }\left( \frac { 5x }{ 3 } \right) +c

Question 5.
\frac { 3x }{ 1+{ 2x }^{ 4 } }
Solution:
Put x²=t,so that 2x dx=dt
⇒x dx = \frac { dt }{ 2 }
\therefore \int { \frac { 3x }{ 1+{ 2x }^{ 4 } } dx } =\frac { 1 }{ 2 } \int { \frac { dt }{ 1+{ 2t }^{ 2 } } } =\frac { 3 }{ 4 } \int { \frac { dt }{ { \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+{ t }^{ 2 } } }
=\frac { 3 }{ 2\sqrt { 2 } } { tan }^{ -1 }(\sqrt { 2t } )+c\quad =\frac { 3 }{ 2\sqrt { 2 } } { tan }^{ -1 }(\sqrt { { 2x }^{ 2 } } )+c

Question 6.
\frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } }
Solution:
put x3 = t,so that 3x²dx = dt
\int { \frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } } dx } \quad =\frac { 1 }{ 3 } \int { \frac { dt }{ 1-{ t }^{ 2 } } \quad =\frac { 1 }{ 6 } log } \left| \frac { 1+t }{ 1-t } \right| +c
=\frac { 1 }{ 6 } log\left| \frac { 1+{ x }^{ 3 } }{ 1-{ x }^{ 3 } } \right| +c

Question 7.
\frac { x-1 }{ \sqrt { { x }^{ 2 }-1 } }
Solution:
I=\int { \frac { x-1 }{ \sqrt { { x }^{ 2 }-1 } } dx } -\int { \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } dx } ,I={ I }_{ 1 }-{ I }_{ 2 }
put x²-1 = t,so that 2x dx = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 7

Question 8.
\frac { { x }^{ 2 } }{ \sqrt { { x }^{ 6 }+{ a }^{ 6 } } }
Solution:
put x3 = t
so that 3x2dx = dt
I=\frac { 1 }{ 3 } \int { \frac { dt }{ { t }^{ 2 }+{ { (a }^{ 3 }) }^{ 2 } } =\frac { 1 }{ 3 } log\left| t+\sqrt { { t }^{ 2 }+{ a }^{ 6 } } \right| +c }
=\frac { 1 }{ 3 } log|{ x }^{ 3 }+\sqrt { { a }^{ 6 }+{ x }^{ 6 } } |+c

Question 9.
\frac { { sec }^{ 2 }x }{ \sqrt { { tan }^{ 2 }x+4 } }
Solution:
let tanx = t
sec x²dx = dt
I=\int { \frac { dt }{ \sqrt { { t }^{ 2 }+{ (2) }^{ 2 } } } } =log|t+\sqrt { { t }^{ 2 }+4 } |+c
=log|tanx+\sqrt { { tan }^{ 2 }x+4 } |+c

Question 10.
\frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } }
Solution:
\int { \frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } } dx } =\int { \frac { dx }{ \sqrt { { (x+1) }^{ 2 }+1 } } }
=log|(x+1)+\sqrt { { x }^{ 2 }+2x+2 } |+c

Question 11.
\frac { 1 }{ { 9x }^{ 2 }+6x+5 }
Solution:
\int { \frac { 1 }{ { 9x }^{ 2 }+6x+5 } } =\frac { 1 }{ 9 } \int { \frac { dx }{ { \left( x+\frac { 1 }{ 3 } \right) }^{ 2 }{ +\left( \frac { 2 }{ 3 } \right) }^{ 2 } } }
=\frac { 1 }{ 6 } { tan }^{ -1 }\left( \frac { 3x+1 }{ 2 } \right) +c

Question 12.
\frac { 1 }{ \sqrt { 7-6x-{ x }^{ 2 } } }
Solution:
I=\int { \frac { dx }{ \sqrt { { 4 }^{ 2 }-{ (x+3) }^{ 2 } } } } \quad ={ sin }^{ -1 }\left( \frac { x+3 }{ 4 } \right) +c

Question 13.
\frac { 1 }{ \sqrt { (x-1)(x-2) } }
Solution:
\int { \frac { 1 }{ \sqrt { (x-1)(x-2) } } dx } =\int { \frac { dx }{ \sqrt { { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } } }
=log\left| x-\frac { 3 }{ 2 } +\sqrt { { x }^{ 2 }-3x+2 } \right| +c

Question 14.
\frac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } } }
Solution:
\int { \frac { dx }{ \sqrt { 8+3x-{ x }^{ 2 } } } } =\int { \frac { dx }{ \sqrt { 8-\left( { x }^{ 2 }-3x \right) } } }
=\int { \frac { dx }{ \sqrt { { \left( \frac { \sqrt { 41 } }{ 2 } \right) }^{ 2 }-{ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 } } } } \quad ={ sin }^{ -1 }\left( \frac { 2x-3 }{ \sqrt { 41 } } \right) +c

Question 15.
\frac { 1 }{ \sqrt { (x-a)(x-b) } }
Solution:
\int { \frac { dx }{ \sqrt { (x-a)(x-b) } } } =\int { \frac { dx }{ { \left( x-\frac { a+b }{ 2 } \right) }^{ 2 }-{ \left( \frac { a-b }{ 2 } \right) }^{ 2 } } }
=log\left| \left( x-\frac { a+b }{ 2 } \right) +\sqrt { (x-a)(x-b) } \right| +c

Question 16.
\frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } }
Solution:
let\quad I=\int { \frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } } } dx
put 2x²+x-3=t
so that (4x+1)dx=dt
let\quad I=\int { \frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } } } dx
\therefore I=\int { \frac { dt }{ \sqrt { t } } } ={ 2t }^{ \frac { 1 }{ 2 } }+c\quad =2\sqrt { { 2x }^{ 2 }+x-3 } +c

Question 17.
\frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } }
Solution:
\int { \frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } } dx } \quad =\int { \frac { x }{ \sqrt { { x }^{ 2 }-1 } } dx } +\int { \frac { 2 }{ \sqrt { { x }^{ 2 }-1 } } dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 17
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 17.1

Question 18.
\frac { 5x-2 }{ 1+2x+{ 3x }^{ 2 } }
Solution:
put 5x-2=A\frac { d }{ dx }(1+2x+3x²)+B
⇒ 6A=5, A=\frac { 5 }{ 6 }-2=2A+B, B=-\frac { 11 }{ 3 }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 18

Question 19.
\frac { 6x+7 }{ \sqrt { (x-5)(x-4) } }
Solution:
\int { \frac { 6x+7 }{ \sqrt { (x-5)(x-4) } } dx } =\int { \frac { (6x+7)dx }{ \sqrt { { x }^{ 2 }-9x+20 } } }
vedantu class 12 maths Chapter 7 Integrals 19
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 19.1

Question 20.
\frac { x+2 }{ \sqrt { 4x-{ x }^{ 2 } } }
Solution:
I=\int { \frac { x-2 }{ \sqrt { 4-{ (x-2) }^{ 2 } } } dx+4\int { \frac { dx }{ \sqrt { 4-{ (x-2) }^{ 2 } } } } }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 20

Question 21.
\frac { x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } }
Solution:
I=\frac { 1 }{ 2 } \int { \frac { 2x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } } dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 21

Question 22.
\frac { x+3 }{ { x }^{ 2 }-2x-5 }
Solution:
I=\frac { 1 }{ 2 } \int { \frac { 2x-2 }{ { x }^{ 2 }-2x-5 } dx } +\int { \frac { dx }{ { x }^{ 2 }-2x-5 } }
vedantu class 12 maths Chapter 7 Integrals 22

Question 23.
\frac { 5x+3 }{ \sqrt { { x }^{ 2 }+4x+10 } }
Solution:
I=\int { \frac { \frac { 5 }{ 2 } (2x+4)+(3-10) }{ \sqrt { { x }^{ 2 }+4x+10 } } dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 23

Question 24.
\int { \frac { dx }{ { x }^{ 2 }+2x+2 } equals }
(a) xtan-1(x+1)+c
(b) (x+1)tan-1x+c
(c) tan-1(x+1)+c
(d) tan-1x+c
Solution:
(b) let\quad I=\int { \frac { dx }{ { x }^{ 2 }+2x+2 } } =\int { \frac { dx }{ (x+1)^{ 2 }+1 } }
= (x+1)tan-1x+c

Question 25.
\int { \frac { dx }{ \sqrt { 9x-{ 4x }^{ 2 } } } equals }
(a) \frac { 1 }{ 9 } { sin }^{ -1 }\left( \frac { 9x-8 }{ 8 } \right) +c
(b) \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 8x-9 }{ 9 } \right) +c
(c) \frac { 1 }{ 3 } { sin }^{ -1 }\left( \frac { 9x-8 }{ 8 } \right) +c
(d) { sin }^{ -1 }\left( \frac { 9x-8 }{ 9 } \right) +c
Solution:
(b) \int { \frac { dx }{ \sqrt { 9x-{ 4x }^{ 2 } } } } =\frac { 1 }{ 2 } \left[ \frac { dx }{ \sqrt { \left( \frac { 9 }{ 8 } \right) ^{ 2 }-\left[ { x }^{ 2 }-{ \frac { 9 }{ 4 } }x+\left( \frac { 9 }{ 8 } \right) ^{ 2 } \right] } } \right]
\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 8x-9 }{ 9 } \right) +c

Exercise 7.5

Integrate the rational function in exercises 1 to 21

Question 1.
\frac { x }{ (x+1)(x+2) }
Solution:
let \frac { x }{ (x+1)(x+2) } ≡ \frac { A }{ x+1 } +\frac { B }{ x+2 }
⇒ x ≡ A(x+2)+B(x+1)….(i)
putting x = -1 & x = -2 in (i)
we get A = 1,B = 2
\therefore \int { \frac { 1 }{ (x+1)(x+2) } dx } =\int { \frac { -1 }{ x+1 } dx } +\int { \frac { 2 }{ x+2 } dx }
=-log|x+1| + 2log|x+2|+c

Question 2.
\frac { 1 }{ { x }^{ 2 }-9 }
Solution:
let \frac { 1 }{ { x }^{ 2 }-9 } =\frac { 1 }{ (x-3)(x+3) } \equiv \frac { A }{ x-3 } +\frac { B }{ x+3 }
⇒ x ≡ A(x+3)+B(x-3)…(i)
put x = 3, -3 in (i)
we get A=\frac { 1 }{ 6 } & B=-\frac { 1 }{ 6 }
\therefore \int { \frac { 1 }{ { x }^{ 2 }-9 } dx } =\frac { 1 }{ 6 } \int { \left[ \frac { 1 }{ x-3 } -\frac { 1 }{ x+3 } \right] dx }
=\frac { 1 }{ 6 } log\left| \frac { x-3 }{ x+3 } \right| +c

Question 3.
\frac { 3x-1 }{ (x-1)(x-2)(x-3) }
Solution:
Let \frac { 3x-1 }{ (x-1)(x-2)(x-3) } =\frac { A }{ x-1 } +\frac { B }{ x-2 } +\frac { C }{ x-3 }
⇒ 3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(-2)…..(i)
put x = 1,2,3 in (i)
we get A = 1,B = -5 & C = 4
\therefore I=\int { \frac { 1 }{ x-1 } dx } -5\int { \frac { 1 }{ x-2 } dx } +4\int { \frac { 1 }{ x-3 } dx }
=log|x-1| – 5log|x-2| + 4log|x+3| + C

Question 4.
\frac { x }{ (x-1)(x-2)(x-3) }
Solution:
let \frac { x }{ (x-1)(x-2)(x-3) } =\frac { A }{ x-1 } +\frac { B }{ x-2 } +\frac { C }{ x-3 }
⇒ x ≡ A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)…(i)
put x = 1,2,3 in (i)
A=\frac { 1 }{ 2 } ,B=-2,C=\frac { 3 }{ 2 }
\therefore I=\frac { 1 }{ 2 } \int { \frac { dx }{ x-1 } } -2\int { \frac { dx }{ x-2 } } +\frac { 3 }{ 2 } \int { \frac { dx }{ x-3 } }
=\frac { 1 }{ 2 } log|x-1|-2log|x-2|+\frac { 3 }{ 2 } log|x-3|+c

Question 5.
\frac { 2x }{ { x }^{ 2 }+3x+2 }
Solution:
let \frac { 2x }{ { x }^{ 2 }+3x+2 } =\frac { 2x }{ (x+1)(x+2) } =\frac { A }{ x+1 } +\frac { B }{ x+2 }
⇒ 2x = A(x+2)+B(x+1)…(i)
put x = -1, -2 in (i)
we get A = -2, B = 4
\therefore \int { \frac { 2x }{ { x }^{ 2 }+3x+2 } dx } =-2\int { \frac { dx }{ x+1 } } +4\int { \frac { dx }{ x+2 } }
=-2log|x+1|+4log|x+2|+c

Question 6.
\frac { 1-{ x }^{ 2 } }{ x(1-2x) }
Solution:
\frac { 1-{ x }^{ 2 } }{ (x-2{ x }^{ 2 }) } is an improper fraction therefore we
convert it into a proper fraction. Divide 1 – x² by x – 2x² by long division.
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 6
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 6.1

Question 7.
\frac { x }{ \left( { x }^{ 2 }+1 \right) \left( x-1 \right) }
Solution:
let \frac { x }{ \left( { x }^{ 2 }+1 \right) \left( x-1 \right) } =\frac { A }{ x-1 } +\frac { Bx+C }{ { x }^{ 2 }+1 }
⇒ x = A(x²+1)+(Bx+C)(x-1)
Put x = 1,0
⇒ A=\frac { 1 }{ 2 } C=\frac { 1 }{ 2 } \Rightarrow B=-\frac { 1 }{ 2 }
\therefore I=\frac { 1 }{ 2 } \int { \frac { dx }{ x-1 } } -\frac { 1 }{ 2 } \int { \frac { x }{ { x }^{ 2 }+1 } dx } +\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+1 } }
=\frac { 1 }{ 2 } log(x-1)-\frac { 1 }{ 4 } log({ x }^{ 2 }+1)+\frac { 1 }{ 2 } { tan }^{ -1 }x+c

Question 8.
\frac { x }{ { \left( x-1 \right) }^{ 2 }\left( x+2 \right) }
Solution:
\frac { x }{ { \left( x-1 \right) }^{ 2 }\left( x+2 \right) } =\frac { A }{ x-1 } +\frac { B }{ { \left( x-1 \right) }^{ 2 } } +\frac { C }{ x+2 }
⇒ x ≡ A(x-1)(x+2)+B(x+2)+C(x-1)² …(i)
put x = 1, -2
we get B=\frac { 1 }{ 3 } ,C=\frac { -2 }{ 9 }
\therefore I=\frac { 2 }{ 9 } \int { \frac { 1 }{ x-1 } dx } +\frac { 1 }{ 3 } \int { \frac { 1 }{ { (x-1) }^{ 2 } } dx } -\frac { 2 }{ 9 } \int { \frac { 1 }{ x+2 } dx }
=\frac { 2 }{ 9 } log\left| \frac { x-1 }{ x+2 } \right| -\frac { 1 }{ 3\left( x-1 \right) } +c

Question 9.
\frac { 3x+5 }{ { x }^{ 3 }-{ x }^{ 2 }-x+1 }
Solution:
let \frac { 3x+5 }{ { x }^{ 2 }(x-1)-1(x-1) }
\frac { 3x+5 }{ (x-1)^{ 2 }(x+1) } =\frac { A }{ x-1 } +\frac { B }{ { (x-1) }^{ 2 } } +\frac { C }{ x+1 }
⇒ 3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)
put x = 1,-1,0
we get B=4,C=\frac { 1 }{ 2 } ,A=-\frac { 1 }{ 2 }
\therefore I=-\frac { 1 }{ 2 } \int { \frac { dx }{ (x-1) } } +4\frac { dx }{ { (x-1) }^{ 2 } } +\frac { 1 }{ 2 } \int { \frac { dx }{ x+1 } }
=\frac { 1 }{ 2 } log\left| \frac { x+1 }{ x-1 } \right| -\frac { 4 }{ x-1 } +c

Question 10.
\frac { 2x-3 }{ ({ x }^{ 2 }-1)(2x+3) }
Solution:
\frac { 2x-3 }{ ({ x }^{ 2 }-1)(2x+3) } =\frac { 2x-3 }{ (x-1)(x+1)(2x+3) }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 10

Question 11.
\frac { 5x }{ (x-1)({ x }^{ 2 }-4) }
Solution:
let \frac { 5x }{ (x-1)({ x }^{ 2 }-4) } =\frac { 5x }{ (x+1)(x+2)(x-2) }
byjus class 12 maths Chapter 7 Integrals 11

Question 12.
\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 }
Solution:
\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 } =x+\frac { 2x+1 }{ (x+1)(x-1) }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 12
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 12.1

Question 13.
\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) }
Solution:
\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } =\frac { A }{ 1-x } +\frac { Bx+C }{ 1+{ x }^{ 2 } }
⇒ 2 = A(1+x²) + (Bx+C)(1 -x) …(i)
Putting x = 1 in (i), we get; A = 1
Also 0 = A – B and 2 = A + C ⇒B = A = 1 & C = 1
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 13

Question 14.
\frac { 3x-1 }{ { (x+2) }^{ 2 } }
Solution:
\frac { 3x-1 }{ { (x+2) }^{ 2 } } \equiv \frac { A }{ x+1 } +\frac { B }{ { (x+2) }^{ 2 } }
=>3x – 1 = A(x + 2) + B …(i)
Comparing coefficients A = -1 and B = -7
\therefore \int { \frac { 3x-1 }{ { (x+2) }^{ 2 } } dx } =3\int { \frac { dx }{ x+2 } } -7\int { \frac { dx }{ { (x+2) }^{ 2 } } }
=3log|x+2|+\frac { 7 }{ x+2 } +c

Question 15.
\frac { 1 }{ { x }^{ 4 }-1 }
Solution:
\frac { 1 }{ { x }^{ 4 }-1 } =\frac { A }{ x+1 } +\frac { B }{ x-1 } +\frac { Cx+D }{ { x }^{ 2 }+1 }
⇒ 1 ≡ A(x-1)(x²+1) + B(x+1)(x²+1) + (Cx+D)(x+1)(x-1) ….(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 15
byjus class 12 maths Chapter 7 Integrals 15.1

Question 16.
\frac { 1 }{ x({ x }^{ n }+1) }
[Hint : multiply numerator and denominator by xn-1 and put xn = t ]
Solution:
\frac { { x }^{ n-1 } }{ x.{ x }^{ n-1 }({ x }^{ n }+1) } =\frac { { x }^{ n-1 } }{ { x }^{ n }({ x }^{ n }+1) }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 16

Question 17.
\frac { cosx }{ (1-sinx)(2-sinx) }
Solution:
put sinx = t
so that cosx dx = dt
\therefore I=\int { \frac { 1 }{ (1-t)(2-t) } dt }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 17

Question 18.
\frac { \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+2 \right) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }+4 \right) }
Solution:
put x²=y
I=1-\frac { 2(2y+5) }{ (y+3)(y+4) }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 18

Question 19.
\frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) }
Solution:
put x²=y
so that 2xdx = dy
\therefore \int { \frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) } dx } =\int { \frac { dy }{ (y+1)(y+3) } }
byjus class 12 maths Chapter 7 Integrals 19

Question 20.
\frac { 1 }{ x({ x }^{ 4 }-1) }
Solution:
put x4 = t
so that 4x3 dx = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 20

Question 21.
\frac { 1 }{ { e }^{ x }-1 }
Solution:
Let ex = t ⇒ ex dx = dt
⇒ dx=\frac { dt }{ t }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 21

Question 22.
choose the correct answer in each of the following :
\int { \frac { xdx }{ (x-1)(x-2) } equals }
(a) log\left| \frac { { (x-1) }^{ 2 } }{ x-2 } \right| +c
(b) log\left| \frac { { (x-2) }^{ 2 } }{ x-1 } \right| +c
(c) log\left| \left( \frac { x-{ 1 }^{ 2 } }{ x-2 } \right) \right| +c
(d) log|(x-1)(x-2)|+c
Solution:
(b) \int { \frac { x }{ (x-1)(x-2) } dx } =\int { \left[ \frac { -1 }{ x-1 } +\frac { 2 }{ x-2 } \right] dx }
log\left| \frac { { (x-2) }^{ 2 } }{ x-1 } \right| +c

Question 23.
\int { \frac { dx }{ x({ x }^{ 2 }+1) } equals }
(a) log|x|-\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c
(b) log|x|+\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c
(c) -log|x|+\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c
(d) \frac { 1 }{ 2 } log|x|+log({ x }^{ 2 }+1)+c
Solution:
(a) let \frac { 1 }{ x\left( { x }^{ 2 }+1 \right) } =\frac { A }{ x } +\frac { Bx+C }{ { x }^{ 2 }+1 }
⇒ 1 = A(x²+1)+(Bx+C)(x)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 22

Exercise 7.6

Integrate the functions in Exercises 1 to 22.

Question 1.
x sinx
Solution:
By part integration
∫x sinx dx = x(-cosx) – ∫1(-cosx)dx
=-x cosx + ∫cosxdx
=-x cosx + sinx + c

Question 2.
x sin3x
Solution:
∫x sin3x dx = x\left( -\frac { cos3x }{ 3 } \right) -\int { 1 } .\left( \frac { -cos3x }{ 3 } \right) dx
=-\frac { 1 }{ 3 } x\quad cos3x+\frac { 1 }{ 9 } sin3x+c

Question 3.
{ x }^{ 2 }{ e }^{ x }
Solution:
\int { { x }^{ 2 }{ e }^{ x } } dx={ x }^{ 2 }{ e }^{ x }-2{ x }{ e }^{ x }+2{ e }^{ x }+c
={ e }^{ x }\left( { x }^{ 2 }-2x+2 \right) +c

Question 4.
x logx
Solution:
\int { xlogx\quad dx } =logx\int { xdx } -\int { \left[ \frac { d }{ dx } (logx)\int { xdx } \right] dx }
=\frac { { x }^{ 2 } }{ 2 } logx-\frac { 1 }{ 2 } \int { x\quad dx } =\frac { { x }^{ 2 } }{ 2 } logx-\frac { 1 }{ 4 } { x }^{ 2 }+c

Question 5.
x log2x
Solution:
\int { x\quad log2xdx } =(log2x)\frac { { x }^{ 2 } }{ 2 } -\int { \frac { 1 }{ 2x } } .2\left( \frac { { x }^{ 2 } }{ 2 } \right) dx
=\frac { { x }^{ 2 } }{ 2 } log|2x|-\frac { 1 }{ 2 } \int { xdx } =\frac { { x }^{ 2 } }{ 2 } log|2x|-\frac { { x }^{ 2 } }{ 4 } +c

Question 6.
{ x }^{ 2 }logx
Solution:
\int { { x }^{ 2 }logxdx } =log|x|\left( \frac { { x }^{ 3 } }{ 3 } \right) -\int { \frac { 1 }{ x } } \left( \frac { { x }^{ 3 } }{ 3 } \right) dx
=\frac { { x }^{ 3 } }{ 3 } log|x|-\frac { 1 }{ 3 } \int { { x }^{ 2 }dx } =\frac { { x }^{ 3 } }{ 3 } log|x|-\frac { { x }^{ 3 } }{ 9 } +c

Question 7.
x\quad { sin }^{ -1 }x
Solution:
I=x\quad { sin }^{ -1 }x.\left( \frac { { x }^{ 2 } }{ 2 } \right) -\int { \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } } .\frac { { x }^{ 2 } }{ 2 } dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 7

Question 8.
x\quad { tan }^{ -1 }x
Solution:
I=x\quad { tan}^{ -1 }x.\left( \frac { { x }^{ 2 } }{ 2 } \right) -\int { \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } .\frac { { x }^{ 2 } }{ 2 } dx
=\frac { { x }^{ 2 } }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 2 } \int { \left( 1-\frac { 1 }{ 1+{ x }^{ 2 } } \right) dx }
=\frac { { x }^{ 2 } }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 2 } x+\frac { 1 }{ 2 } { tan }^{ -1 }x+c

Question 9.
x\quad { cos }^{ -1 }x
Solution:
let I = \int { x } { cos }^{ -1 }xdx=\int { { cos }^{ -1 }x } .xdx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 9

Question 10.
{ (sin }^{ -1 }{ x })^{ 2 }
Solution:
put\quad { sin }^{ -1 }x=\theta \Rightarrow x=sin\theta \Rightarrow dx=cos\theta d\theta
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 10

Question 11.
\frac { x\quad { cos }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } }
Solution:
put\quad { cos }^{ -1 }x=t\quad so\quad that\frac { x\quad { cos }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } dx=dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 11

Question 12.
x sec²x
Solution:
∫x sec²x dx =x(tanx)-∫1.tanx dx
= x tanx+log cosx+c

Question 13.
{ ta }n^{ -1 }x
Solution:
\int { { tan }^{ -1 }xdx } =x{ tan }^{ -1 }x-\frac { 1 }{ 2 } \int { \frac { 2x }{ 1+{ x }^{ 2 } } dx }
=x{ tan }^{ -1 }x-\frac { 1 }{ 2 } log|1+{ x }^{ 2 }|+c

Question 14.
x(logx)²
Solution:
∫x(logx)² dx
=\frac { { x }^{ 2 } }{ 2 } { (logx) }^{ 2 }-\left[ (logx)\frac { { x }^{ 2 } }{ 2 } -\int { \frac { 1 }{ x } \frac { { x }^{ 2 } }{ 2 } dx } \right]
=\frac { { x }^{ 2 } }{ 2 } { (logx) }^{ 2 }-\frac { { x }^{ 2 } }{ 2 } logx+\frac { 1 }{ 4 } { x }^{ 2 }+c

Question 15.
(x²+1)logx
Solution:
∫(x²+1)logx dx
=logx\left( \frac { { x }^{ 3 } }{ 3 } +x \right) -\int { \frac { 1 }{ x } \left( \frac { { x }^{ 3 } }{ 3 } +x \right) dx }
=\left( \frac { { x }^{ 3 } }{ 3 } +x \right) logx-\frac { { x }^{ 3 } }{ 9 } -x+c

Question 16.
{ e }^{ x }(sinx+cosx)
Solution:
put\quad { e }^{ x }sinx=t\Rightarrow { e }^{ x }(sinx+cosx)dx=dt
\therefore \int { { e }^{ x }(sinx+cosx)dx } =\int { dt } =t+c
={ e }^{ x }sinx+c

Question 17.
\frac { { xe }^{ x } }{ { (1+x) }^{ 2 } }
Solution:
\int { \frac { { xe }^{ x } }{ { (1+x) }^{ 2 } } }
tiwari academy class 12 maths Chapter 7 Integrals 17

Question 18.
\frac { { e }^{ x }(1+sinx) }{ 1+cosx }
Solution:
I=\int { { e }^{ x } } \left[ \frac { 1+2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ 2{ cos }^{ 2 }\frac { x }{ 2 } } \right] dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 18

Question 19.
{ e }^{ x }\left( \frac { 1 }{ x } -\frac { 1 }{ { x }^{ 2 } } \right)
Solution:
put \frac { { e }^{ x } }{ x } =t\Rightarrow { e }^{ x }\left( \frac { 1 }{ x } -\frac { 1 }{ { x }^{ 2 } } \right) dx=dt
\therefore I=\int { dt } =t+c=\frac { { e }^{ x } }{ x } +c

Question 20.
\frac { { (x-2)e }^{ x } }{ { (x-1) }^{ 3 } }
Solution:
I=\int { { e }^{ x }\left[ \frac { 1 }{ { (x-1) }^{ 2 } } -\frac { 2 }{ { (x-1) }^{ 3 } } \right] dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 19

Question 21.
{ e }^{ 2x }sinx
Solution:
let I=\int { { e }^{ 2x }sinx }
={ e }^{ 2x }(-cosx)-\int { 2{ e }^{ 2x }(-cosx)dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 21

Question 22.
{ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right)
Solution:
Put x = tan t
so that dx = sec² t dt
tiwari academy class 12 maths Chapter 7 Integrals 22

Choose the correct answer in exercise 23 and 24

Question 23.
\int { { x }^{ 2 }{ e }^{ { x }^{ 3 } } } dx\quad equals
(a) \frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+c
(b) \frac { 1 }{ 3 } +{ e }^{ { x }^{ 2 } }+c
(c) \frac { 1 }{ 2 } { e }^{ { x }^{ 3 } }+c
(d) \frac { 1 }{ 2 } { e }^{ { x }^{ 2 } }+c
Solution:
(a) let x³ = t
⇒3x² dx = dt
\therefore \int { { x }^{ 2 }{ e }^{ { x }^{ 3 } }dx } =\frac { 1 }{ 3 } \int { { e }^{ t }dt } =\frac { 1 }{ 3 } { e }^{ t }+c=\frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+c

Question 24.
\int { { e }^{ x }secx(1+tanx) } dx\quad equals
(a) { e }^{ x }cosx+c
(b) { e }^{ x }secx+c
(c) { e }^{ x }sinx+c
(d) { e }^{ x }tanx+c
Solution:
(b) \int { { e }^{ x }(secx+secx\quad tanx)dx } ={ e }^{ x }secx+c

Exercise 7.7

Integral the function in exercises 1 to 9

Question 1.
\sqrt { 4-{ x }^{ 2 } }
Solution:
let\quad I=\int { \sqrt { 4-{ x }^{ 2 } } } dx=\int { \sqrt { { (2) }^{ 2 }-{ x }^{ 2 } } dx }
=\frac { x\sqrt { 4-{ x }^{ 2 } } }{ 2 } +2{ sin }^{ -1 }\left( \frac { x }{ 2 } \right) +c

Question 2.
\sqrt { 1-{ 4x }^{ 2 } }
Solution:
\int { \sqrt { 1-{ 4x }^{ 2 } } } dx=2\int { \sqrt { { \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ x }^{ 2 } } } dx
=\frac { x\sqrt { 1-{ 4x }^{ 2 } } }{ 2 } +\frac { 1 }{ 4 } { sin }^{ -1 }(2x)+c

Question 3.
\sqrt { { x }^{ 2 }+4x+6 }
Solution:
\int { \sqrt { { x }^{ 2 }+4x+6 } } dx=\int { \sqrt { { (x+2) }^{ 2 }+{ (\sqrt { 2 } ) }^{ 2 } } } dx
=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x+6 } +log\left| (x+2)+\sqrt { { x }^{ 2 }+4x+6 } \right| +c

Question 4.
\sqrt { { x }^{ 2 }+4x+1 }
Solution:
\int { \sqrt { { x }^{ 2 }+4x+1 } } dx=\int { \sqrt { { (x+2) }^{ 2 }-{ (\sqrt { 3 } ) }^{ 2 } } } dx
=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x+1 } -\frac { 3 }{ 2 } log\left| (x+2)+\sqrt { { x }^{ 2 }+4x+1 } \right| +c

Question 5.
\sqrt { 1-4x-{ x }^{ 2 } }
Solution:
\int { \sqrt { 1-4x-{ x }^{ 2 } } } dx=\int { \sqrt { { (5) }^{ 2 }-{ (x+2) }^{ 2 } } dx }
=\frac { x+2 }{ 2 } \sqrt { 5-{ (x+2) }^{ 2 } } dx

Question 6.
\sqrt { { x }^{ 2 }+4x-5 }
Solution:
\int { \sqrt { { x }^{ 2 }+4x-5 } } dx=\int { \sqrt { { (x+2) }^{ 2 }-{ (3) }^{ 2 } } } dx
=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x-5 } -\frac { 9 }{ 2 } log|x+2+\sqrt { { x }^{ 2 }+4x-5 } |+c

Question 7.
\sqrt { 1+3x-{ x }^{ 2 } }
Solution:
\int { \sqrt { 1-\left( { x }^{ 2 }-3x \right) } } dx
=\int { \sqrt { { \left( \frac { \sqrt { 13 } }{ 2 } \right) }^{ 2 }-{ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 } } } dx
=\frac { 2x-3 }{ 4 } \sqrt { 1+3x-{ x }^{ 2 } } +\frac { 13 }{ 8 } { sin }^{ -1 }\left[ \frac { 2x-3 }{ \sqrt { 3 } } \right] +c

Question 8.
\sqrt { { x }^{ 2 }+3x }
Solution:
\int { \sqrt { { x }^{ 2 }+3x } } dx=\int { \sqrt { { \left( x+\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 } } } dx
=\frac { 2x+3 }{ 4 } \sqrt { { x }^{ 2 }+3x } -\frac { 9 }{ 8 } log\left| x+\frac { 3 }{ 2 } +\sqrt { { x }^{ 2 }+3x } \right| +c

Question 9.
\sqrt { 1+\frac { { x }^{ 2 } }{ 9 } }
Solution:
\int { \sqrt { 1+\frac { { x }^{ 2 } }{ 9 } } } dx=\frac { 1 }{ 3 } \int { \sqrt { { x }^{ 2 }+{ 3 }^{ 2 } } }
=\frac { 1 }{ 6 } \left[ x\sqrt { { x }^{ 2 }+9 } +9log|x+\sqrt { { x }^{ 2 }+9 } | \right] +c

Choose the correct answer in the Exercises 10 to 11:

Question 10.
\int { \sqrt { 1+{ x }^{ 2 } } } dx\quad is\quad equal\quad to
(a) \frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log|x+\sqrt { 1+{ x }^{ 2 } } |+c
(b) \frac { 2 }{ 3 } { \left( 1+{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+c
(c) \frac { 2 }{ 3 } x{ \left( 1+{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+c
(d) \frac { { x }^{ 2 } }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } { x }^{ 2 }log\left| x+\sqrt { 1+{ x }^{ 2 } } \right| +c
Solution:
(a) \int { \sqrt { 1+{ x }^{ 2 } } } dx
=\frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log(x+\sqrt { 1+{ x }^{ 2 } } )+c

Question 11.
\int { \sqrt { { x }^{ 2 }-8x+7 } } dx\quad is\quad equal\quad to
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 11
Solution:
(d) \int { \sqrt { { x }^{ 2 }-8x+7 } } dx=\int { \sqrt { { (x-4) }^{ 2 }-{ (3) }^{ 2 } } } dx
=\frac { x-4 }{ 2 } \sqrt { { x }^{ 2 }-8x+7 } -\frac { 9 }{ 2 } log\left| (x-4)+\sqrt { { x }^{ 2 }+8x+7 } \right| +c


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