These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.5
Question 1.
\(\frac { x }{ (x+1)(x+2) }\)
Solution:
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
\(\frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2}\)
Integrating w.r.t. x, we get
\(\int \frac{x}{(x+1)(x+2)} d x=\int \frac{-1}{x+1} d x+\int \frac{2}{x+2} d x\)
\(\frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2}\)
= – log|x + 1| + 2log|x + 2| + C
= – log|x + 1| + log(x + 2)² + C
= log\(\frac{(x+2)^{2}}{|x+1|}\) + C
Question 2.
\(\frac { 1 }{ { x }^{ 2 }-9 } \)
Solution:
Let
\(\frac { 1 }{ { x }^{ 2 }-9 } =\frac { 1 }{ (x-3)(x+3) } \equiv \frac { A }{ x-3 } +\frac { B }{ x+3 } \)
⇒ 1 = A(x + 3) + B(x – 3) … (i)
Put x = 3 in (1), we get 1 = 6B ∴ B = \(\frac { 1 }{ 6 }\)
Put x = – 3 in (1), we get 1 = – 6A ∴ A = \(\frac { – 1 }{ 6 }\)
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 in (1), we get 2 = 2A ∴ A = 1
Put x = 2 in (1), we get 5 = – B ∴ B = – 5
Put x = 3 in (1), we get 8 = 2C ∴ C = 4
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 in (1), we get 1 = 2A ∴ A = \(\frac { 1 }{ 2 }\)
Put x = 2 in (1), we get 2 = – B ∴ B = – 2
Put x = 3 in (1), we get 3 = 2C ∴ C = \(\frac { 3 }{ 2 }\)
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
\(\int \frac{2 x}{(x+1)(x+2)} d x=-2 \int \frac{1}{x+1} d x+4 \int \frac{1}{x+2} d x\)
= – 2log|x + 1| + 4log|x + 2| + C
= 4log|x + 2| – 2log|x + 1| + C
Question 6.
\(\frac { 1-{ x }^{ 2 } }{ x(1-2x) } \)
Solution:
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) … (1)
Put x = 1 in (1), we get 1 = 2A
∴ A = \(\frac { 1 }{ 2 }\)
Equating the coefficients of x² and constant term, we get A + B = 0 and A – C = 0
∴ B = \(\frac { 1 }{ 2 }\) and C = \(\frac { 1 }{ 2 }\)
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 in (1), we get 1 = 3B ∴ B = \(\frac { 1 }{ 3 }\)
Put x = – 2 in (1), we get – 2 = 9C ∴ C = \(\frac { – 2 }{ 9 }\)
Equating the coefficients of x², we get A + C = 0
Hence A = \(\frac { 2 }{ 9 }\)
Question 9.
\(\frac { 3x+5 }{ { x }^{ 3 }-{ x }^{ 2 }-x+1 } \)
Solution:
x³ – x² – x + 1 = x²(x – 1) – 1(x – 1)
= (x² – 1)(x – 1)
= (x + 1) (x – 1) (x – 1)
= (x + 1)(x – 1)²
∴ \(\frac { 3x+5 }{ { x }^{ 3 }-{ x }^{ 2 }-x+1 } \) = \(\frac{3 x+5}{(x+1)(x-1)^{2}}\)
Let \(\frac{3 x+5}{(x+1)(x-1)^{2}}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{(x-1)}+\frac{\mathrm{C}}{(x-1)^{2}}\)
3x + 5 = A(x – 1) + B(x – 1)(x + 1) + C (x + 1) … (1)
Put x = – 1 in (1), we get 2 = 4A ∴ A = \(\frac { 1 }{ 2 }\)
Put x = 1 in (1), we get 8 = 2C ∴ C = 4
Equating the coefficients of x², we get A + B = 0
∴ B = \(\frac { -1 }{ 2 }\)
Question 10.
\(\frac { 2x-3 }{ ({ x }^{ 2 }-1)(2x+3) } \)
Solution:
Let \(\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+1}+\frac{\mathrm{C}}{2 x+3}\)
2x – 3 = A(x + 1)(2x + 3) + B(x – 1)(2x + 3) + C(x – 1)(x + 1) … (1)
Put x = 1 in (1), we get – 1 = 10A ∴ A = \(\frac { – 1 }{ 10 }\)
Put x = – 1 in (1), we get – 5 = – 2B ∴ B = \(\frac { 5 }{ 2 }\)
Put x = \(\frac { – 3 }{ 2 }\) in (1), we get – 6 = \(\frac { 5 }{ 4 }\)C ∴ C = \(\frac { -24 }{ 5 }\)
Question 11.
\(\frac { 5x }{ (x-1)({ x }^{ 2 }-4) } \)
Solution:
\(\frac { 5x }{ (x-1)({ x }^{ 2 }-4) } =\frac { 5x }{ (x+1)(x+2)(x-2) } \)
Let
\(\frac { 5x }{ (x+1)(x+2)(x-2) }\) = \(\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C}}{x-2}\)
5x = A(x + 2) (x – 2) + B(x + 1) (x – 2) + C (x + 1) (x + 2) … (1)
Put x = – 1 in (1), we get – 5 = – 3 A ∴ A = \(\frac { 5 }{ 3 }\)
Put x = – 2 in (1), we get – 10 = 4B ∴ B = \(\frac { – 5 }{ 2 }\)
Put x = 2 in (1), we get 10 = 12C ∴ C = \(\frac { 5 }{ 6 }\)
Question 12.
\(\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 } \)
Solution:
Degree of Nr > Degree of Dr
Question 13.
\(\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } \)
Solution:
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)
Put x = – 2 in (1), we get – 7 = B ∴B = – 7
Equating the coefficients of x, we get A = 3
∴ \(\frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}\)
\(\int \frac{3 x-1}{(x+2)^{2}} d x=3 \int \frac{1}{(x+2)} d x-7 \int \frac{1}{(x+2)^{2}} d x\)
= 3 log|x+2| – 7(\(\left(\frac{-1}{x+2}\right)\)) + C
= \(3log|x+2|+\frac { 7 }{ x+2 } +C\)
Question 15.
\(\frac { 1 }{ { x }^{ 4 }-1 } \)
Solution:
\(\frac { 1 }{ { x }^{ 4 }-1 } \)
= \(\frac{1}{\left(x^{2}-1\right)\left(x^{2}+1\right)}\)
= \(\overline{(x-1)(x+1)\left(x^{2}+1\right)}\)
Let
\(\frac { 1 }{ { x }^{ 4 }-1 } =\frac { A }{ x+1 } +\frac { B }{ x-1 } +\frac { Cx+D }{ { x }^{ 2 }+1 } \)
1 = A(x + 1)(x² + 1) + B(x – 1)(x² + 1) + (Cx + D)(x² – 1)
1 = A(x + 1)(x² + 1) + B(x – 1)(x² + 1) + C(x³ – x) + D(x² – 1) … (1)
Put x = 1 in (1), we get 1 = 4A
∴ A = \(\frac { 1 }{ 4 }\)
Put x = -1 in (1), we get 1 = – 4B
∴ B = \(\frac { – 1 }{ 4 }\)
Equating coefficients of x³ and constant terms, we get,
A + B + C = 0
A – B – D = 1
∴ C = 0 and D = \(\frac { – 1 }{ 2 }\)
\(\frac{1}{x^{4}-1}=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2\left(x^{2}+1\right)}\)
Integrating w.r.t. x, we get
Question 16.
\(\frac { 1 }{ x({ x }^{ n }+1) } \)
Solution
Multiply numerator and denominator by xn-1
Question 17.
\(\frac { cosx }{ (1-sinx)(2-sinx) } \)
Solution:
Put t = sinx, \(\frac { dt }{ dx }\) = cosx, dt = cosx dx
Question 18.
\(\frac { \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+2 \right) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }+4 \right) } \)
Solution:
Question 19.
\(\frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) } \)
Solution:
put x² = y
so that 2x dx = dy
\(\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x=\int \frac{d t}{(t+1)(t+3)}\)
Let \(\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}\)
∴ 1 = A(t + 3) + B(t + 1) … (1)
Put t = – 1 in (1), we get 1 = 2A ∴ A = \(\frac { 1 }{ 2 }\)
Put x = – 3 in (1), we get 1 = – 2B ∴ B = \(\frac { – 1 }{ 2 }\)
Question 20.
\(\frac { 1 }{ x({ x }^{ 4 }-1) } \)
Solution:
Question 21.
\(\frac { 1 }{ { e }^{ x }-1 } \)
Solution:
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:
Let \(\frac{x}{(x-1)(x-2)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}\)
∴ x = A(x – 2) + B(x – 1) … (1)
Put x = 1 in (1), we get 1 = – A ∴ A = – 1
Put x = 2 in (1), we get B = 1
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:
\(\int { \frac { dx }{ x({ x }^{ 2 }+1) } } \) dx
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
Equating the coefficients of x², x and con-stant term, we get A + B = 0, C = 0, A = 1
Solving the equations, we get A = 1, B = – 1, C = 0
\(\frac{1}{x\left(x^{2}+1\right)}=\frac{1}{x}+\frac{-x}{x^{2}+1}\)
\(\int \frac{1}{x\left(x^{2}+1\right)} d x=\int \frac{1}{x} d x-\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x\)
= \(\log |x|-\frac{1}{2} \log \left|x^{2}+1\right|\) + C