NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 7
Chapter Name	Integrals
Exercise	Ex 7.5
Number of Questions Solved	23
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 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

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 $\frac { 5x }{ (x-1)({ x }^{ 2 }-4) } =\frac { 5x }{ (x+1)(x+2)(x-2) }$
NCERT Solutions for 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

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

Question 16.
$\frac { 1 }{ x({ x }^{ n }+1) }$
[Hint : multiply numerator and denominator by x^n-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
$\therefore \int { \frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) } dx } =\int { \frac { dy }{ (y+1)(y+3) } }$
NCERT Solutions for 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 e^x = t ⇒ e^x 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

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5

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