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.

• Integrals Class 12 Ex 7.1
• Integrals Class 12 Ex 7.2
• Integrals Class 12 Ex 7.3
• Integrals Class 12 Ex 7.5
• Integrals Class 12 Ex 7.6
• Integrals Class 12 Ex 7.7
• Integrals Class 12 Ex 7.8
• Integrals Class 12 Ex 7.9
• Integrals Class 12 Ex 7.10
• Integrals Class 12 Ex 7.11

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

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:
\(\int { \frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } } dx } \quad =\int { \frac { x }{ \sqrt { { x }^{ 2 }-1 } } dx } +\int { \frac { 2 }{ \sqrt { { x }^{ 2 }-1 } } dx } \)

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=\(-\frac { 11 }{ 3 }\)

Ex 7.4 Class 12 Maths Question 19.
\(\frac { 6x+7 }{ \sqrt { (x-5)(x-4) } } \)
Solution:
\(\int { \frac { 6x+7 }{ \sqrt { (x-5)(x-4) } } dx } =\int { \frac { (6x+7)dx }{ \sqrt { { x }^{ 2 }-9x+20 } } } \)

Ex 7.4 Class 12 Maths Question 20.
\(\frac { x+2 }{ \sqrt { 4x-{ x }^{ 2 } } } \)
Solution:
\(I=\int { \frac { x-2 }{ \sqrt { 4-{ (x-2) }^{ 2 } } } dx+4\int { \frac { dx }{ \sqrt { 4-{ (x-2) }^{ 2 } } } } } \)

Ex 7.4 Class 12 Maths Question 21.
\(\frac { x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } } \)
Solution:
\(I=\frac { 1 }{ 2 } \int { \frac { 2x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } } dx } \)

Ex 7.4 Class 12 Maths Question 22.
\(\frac { x+3 }{ { x }^{ 2 }-2x-5 } \)
Solution:
\(I=\frac { 1 }{ 2 } \int { \frac { 2x-2 }{ { x }^{ 2 }-2x-5 } dx } +\int { \frac { dx }{ { x }^{ 2 }-2x-5 } } \)

Ex 7.4 Class 12 Maths Question 23.
\(\frac { 5x+3 }{ \sqrt { { x }^{ 2 }+4x+10 } } \)
Solution:
\(I=\int { \frac { \frac { 5 }{ 2 } (2x+4)+(3-10) }{ \sqrt { { x }^{ 2 }+4x+10 } } dx } \)

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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