NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 7
Chapter Name	Integrals
Exercise	Ex 7.4
Number of Questions Solved	25
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Integrate the functions in exercises 1 to 23

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths Question 2.
$\frac { 1 }{ \sqrt { 1+{ 4x }^{ 2 } } }$
Solution:
$\frac { 1 }{ 2 } \int { \frac { dx }{ \sqrt { \frac { 1 }{ 4 } +{ x }^{ 2 } } } } =\frac { 1 }{ 2 } \int { \frac { dx }{ \sqrt { { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ x }^{ 2 } } } }$
$=\frac { 1 }{ 2 } log\left| 2x+\sqrt { 1+{ 4x }^{ 2 } } \right| +c$

Ex 7.4 Class 12 Maths Question 3.
$\frac { 1 }{ \sqrt { { (2-x) }^{ 2 }+1 } }$
Solution:
put (2-x)=t
so that -dx=dt
⇒ dx=-dt
$\int { \frac { dx }{ \sqrt { { (2-x) }^{ 2 }+1 } } } =-\int { \frac { dt }{ \sqrt { { t }^{ 2 }+1 } } } =-log|t+\sqrt { { t }^{ 2 }+1 } |+c$
$=log\left| \frac { 1 }{ (2-x)+\sqrt { { x }^{ 2 }-4x+5 } } \right| +c$

Ex 7.4 Class 12 Maths Question 4.
$\frac { 1 }{ \sqrt { 9-{ 25x }^{ 2 } } }$
Solution:
$\int { \frac { dx }{ \sqrt { 9-{ 25x }^{ 2 } } } } =\frac { 1 }{ 5 } \int { \frac { dx }{ \sqrt { { \left( \frac { 3 }{ 5 } \right) }^{ 2 }-{ x }^{ 2 } } } }$
$=\frac { 1 }{ 5 } { sin }^{ -1 }\left( \frac { x }{ \frac { 3 }{ 5 } } \right) +c\quad =\frac { 1 }{ 5 } { sin }^{ -1 }\left( \frac { 5x }{ 3 } \right) +c$

Ex 7.4 Class 12 Maths Question 5.
$\frac { 3x }{ 1+{ 2x }^{ 4 } }$
Solution:
Put x²=t,so that 2x dx=dt
⇒x dx = $\frac { dt }{ 2 }$
$\therefore \int { \frac { 3x }{ 1+{ 2x }^{ 4 } } dx } =\frac { 1 }{ 2 } \int { \frac { dt }{ 1+{ 2t }^{ 2 } } } =\frac { 3 }{ 4 } \int { \frac { dt }{ { \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+{ t }^{ 2 } } }$
$=\frac { 3 }{ 2\sqrt { 2 } } { tan }^{ -1 }(\sqrt { 2t } )+c\quad =\frac { 3 }{ 2\sqrt { 2 } } { tan }^{ -1 }(\sqrt { { 2x }^{ 2 } } )+c$

Ex 7.4 Class 12 Maths Question 6.
$\frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } }$
Solution:
put x³ = t,so that 3x²dx = dt
$\int { \frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } } dx } \quad =\frac { 1 }{ 3 } \int { \frac { dt }{ 1-{ t }^{ 2 } } \quad =\frac { 1 }{ 6 } log } \left| \frac { 1+t }{ 1-t } \right| +c$
$=\frac { 1 }{ 6 } log\left| \frac { 1+{ x }^{ 3 } }{ 1-{ x }^{ 3 } } \right| +c$

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths Question 8.
$\frac { { x }^{ 2 } }{ \sqrt { { x }^{ 6 }+{ a }^{ 6 } } }$
Solution:
put x³ = t
so that 3x²dx = dt
$I=\frac { 1 }{ 3 } \int { \frac { dt }{ { t }^{ 2 }+{ { (a }^{ 3 }) }^{ 2 } } =\frac { 1 }{ 3 } log\left| t+\sqrt { { t }^{ 2 }+{ a }^{ 6 } } \right| +c }$
$=\frac { 1 }{ 3 } log|{ x }^{ 3 }+\sqrt { { a }^{ 6 }+{ x }^{ 6 } } |+c$

Ex 7.4 Class 12 Maths Question 9.
$\frac { { sec }^{ 2 }x }{ \sqrt { { tan }^{ 2 }x+4 } }$
Solution:
let tanx = t
sec x²dx = dt
$I=\int { \frac { dt }{ \sqrt { { t }^{ 2 }+{ (2) }^{ 2 } } } } =log|t+\sqrt { { t }^{ 2 }+4 } |+c$
$=log|tanx+\sqrt { { tan }^{ 2 }x+4 } |+c$

Ex 7.4 Class 12 Maths Question 10.
$\frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } }$
Solution:
$\int { \frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } } dx } =\int { \frac { dx }{ \sqrt { { (x+1) }^{ 2 }+1 } } }$
$=log|(x+1)+\sqrt { { x }^{ 2 }+2x+2 } |+c$

Ex 7.4 Class 12 Maths Question 11.
$\frac { 1 }{ { 9x }^{ 2 }+6x+5 }$
Solution:
$\int { \frac { 1 }{ { 9x }^{ 2 }+6x+5 } } =\frac { 1 }{ 9 } \int { \frac { dx }{ { \left( x+\frac { 1 }{ 3 } \right) }^{ 2 }{ +\left( \frac { 2 }{ 3 } \right) }^{ 2 } } }$
$=\frac { 1 }{ 6 } { tan }^{ -1 }\left( \frac { 3x+1 }{ 2 } \right) +c$

Ex 7.4 Class 12 Maths Question 12.
$\frac { 1 }{ \sqrt { 7-6x-{ x }^{ 2 } } }$
Solution:
$I=\int { \frac { dx }{ \sqrt { { 4 }^{ 2 }-{ (x+3) }^{ 2 } } } } \quad ={ sin }^{ -1 }\left( \frac { x+3 }{ 4 } \right) +c$

Ex 7.4 Class 12 Maths Question 13.
$\frac { 1 }{ \sqrt { (x-1)(x-2) } }$
Solution:
$\int { \frac { 1 }{ \sqrt { (x-1)(x-2) } } dx } =\int { \frac { dx }{ \sqrt { { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } } }$
$=log\left| x-\frac { 3 }{ 2 } +\sqrt { { x }^{ 2 }-3x+2 } \right| +c$

Ex 7.4 Class 12 Maths Question 14.
$\frac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } } }$
Solution:
$\int { \frac { dx }{ \sqrt { 8+3x-{ x }^{ 2 } } } } =\int { \frac { dx }{ \sqrt { 8-\left( { x }^{ 2 }-3x \right) } } }$
$=\int { \frac { dx }{ \sqrt { { \left( \frac { \sqrt { 41 } }{ 2 } \right) }^{ 2 }-{ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 } } } } \quad ={ sin }^{ -1 }\left( \frac { 2x-3 }{ \sqrt { 41 } } \right) +c$

Ex 7.4 Class 12 Maths Question 15.
$\frac { 1 }{ \sqrt { (x-a)(x-b) } }$
Solution:
$\int { \frac { dx }{ \sqrt { (x-a)(x-b) } } } =\int { \frac { dx }{ { \left( x-\frac { a+b }{ 2 } \right) }^{ 2 }-{ \left( \frac { a-b }{ 2 } \right) }^{ 2 } } }$
$=log\left| \left( x-\frac { a+b }{ 2 } \right) +\sqrt { (x-a)(x-b) } \right| +c$

Ex 7.4 Class 12 Maths 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
$let\quad I=\int { \frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } } } dx$
$\therefore I=\int { \frac { dt }{ \sqrt { t } } } ={ 2t }^{ \frac { 1 }{ 2 } }+c\quad =2\sqrt { { 2x }^{ 2 }+x-3 } +c$

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths 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

Ex 7.4 Class 12 Maths 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) $\int { \frac { dx }{ \sqrt { 9x-{ 4x }^{ 2 } } } } =\frac { 1 }{ 2 } \left[ \frac { dx }{ \sqrt { \left( \frac { 9 }{ 8 } \right) ^{ 2 }-\left[ { x }^{ 2 }-{ \frac { 9 }{ 4 } }x+\left( \frac { 9 }{ 8 } \right) ^{ 2 } \right] } } \right]$
$\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 8x-9 }{ 9 } \right) +c$

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

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