GSEB Solutions Class 12 Maths Chapter 7 Integrals Ex 7.5

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 7 Integrals Ex 7.5 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Integrals Ex 7.5

Integrate the following functions:
Question 1.
$\frac{x}{(x+1)(x+2)}$
Solution:
$\frac{x}{(x+1)(x+2)}$ may be written as
$\frac{x}{(x+1)(x+2)}$ = $\frac{A}{x+1}$ + $\frac{B}{x+2}$
∴ x = A(x + 2) + B(x + 1)
Put x = – 1, – 1 = A(- 1 + 2) + 0 = A ∴ A = – 1.
Put x = – 2, – 2 = 0 + B(- 2 + 1) = – B ∴ B = 2.
∴ $\frac{x}{(x+1)(x+2)}$ = – $\frac{1}{x+1}$ + $\frac{2}{x+2}$.
Integrating both sides, we get

Question 2.
$\frac{1}{x^{2}-9}$
Solution:
Let $\frac{1}{x^{2}-9}$ = $\frac{1}{(x-3)(x+3)}$ = $\frac{A}{x-3}$ + $\frac{B}{x+3}$.
⇒ 1 = A(x + 3) + B(x – 3) ……………… (1)
Putting x = 3 in (1), we get:
1 = A(3 + 3) ⇒ A = $\frac{1}{6}$.
Putting x = – 3 in (1), we get:
1 = B(- 3 – 3) ⇒ B = – $\frac{1}{6}$

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)(x – 2) ……………….. (1)
Putting x = 1 in (1) we get:
3 – 1 = A(1 – 2)(1 – 3)
⇒ 2 = A(- 1)(- 2) ⇒ A = 1.
Putting x = 2 in (1), we get:
6 – 1 = B(2 – 1)(2 – 3)
⇒ 5 = B(1)(- 1) ⇒ B = – 5.
Putting x = 3 in (1), we get:
9 – 1 = C(3 – 1)(3 – 2)
⇒ 8 = C(2)(1) ⇒ C = 4.

= log |x – 1| – 5 log |x – 2| + 4 log |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) …………. (1)
Putting x = 1 in (1), we get:
1 = A(1 – 2)(1 – 3)
⇒ A = $\frac{1}{2}$
Putting x = 2 in (1), we get:
2 = B(2 – 1)(2 – 3) ⇒ B = – 2
Putting x = 3 in (1), we get:
3 = C(3 – 1)(3 – 2) ⇒ C = $\frac{3}{2}$.

Question 5.
$\frac{2 x}{x^{2}+3 x+2}$
Solution:

Question 6.
$\frac{1-x^{2}}{x(1-2 x)}$
Solution:
Since $\frac{1-x^{2}}{x(1-2 x)}$ = $\frac{1-x^{2}}{x-2 x^{2}}$ is an improper function,
therefore we convert it into a proper function. Divide 1 – x² by x – 2x² by long division method, we get

⇒ x – 2 = A(2x – 1) + Bx ………………. (2)
Puttting x = 0 in (2), we get:
– 2 = A(- 1) ⇒ A = 2.
Putting x = $\frac{1}{2}$ in (2), we get:

Question 7.
$\frac{x}{\left(x^{2}+1\right)(x-1)}$
Solution:
$\frac{x}{\left(x^{2}+1\right)(x-1)}$ may be written as
$\frac{x}{\left(x^{2}+1\right)(x-1)}$ = $\frac{A}{x-1}$ + $\frac{B x+C}{x^{2}-1}$.
∴ x = A(x² + 1) + (Bx + C)(x – 1)
= A(x² + 1) + B(x² – x) + C(x – 1)
Put x = 1, 1 = A(1 + 1) ⇒ 1 = 2A or A = $\frac{1}{2}$
Comparing the coefficients of x², we get
0 = A + B
∴ B = – A = – $\frac{1}{2}$.
Comparing the constants, we get
0 = A – C ∴ C = A = $\frac{1}{2}$.

Question 8.
$\frac{x}{(x-1)^{2}(x+2)}$
Solution:
Let $\frac{x}{(x-1)^{2}(x+2)}$ = $\frac{A}{x-1}$ + $\frac{B}{(x-1)^{2}}$ + $\frac{C}{x+2}$.
⇒ x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² …………….. (1)
Putting x = 1 in (1), we get:
1 = B(1 + 2) ⇒ B = $\frac{1}{3}$
Putting x = – 2 in (1), we get:
– 2 = C(- 2 – 1)² ⇒ C = – $\frac{-2}{9}$
Comparing co-efficients of x² on both sides of (1), we have:
0 = A + C ⇒ A – C = $\frac{2}{9}$.

Question 9.
$\frac{3 x+5}{x^{3}-x^{2}-x+1}$
Solution:

Question 10.
$\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}$
Solution:
Let $\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}$ = $\frac{2x-3}{(x-1)(x+1)(2x+3)}$
= $\frac{A}{x-1}$ + $\frac{B}{x+1}$ + $\frac{C}{2x+3}$
⇒ 2x – 3 = A(x + 1)(2x + 3) + B(x – 1)(2x + 3) + C(x – 1)(x + 1) …………….. (1)
Putting x = 1 in (1), we get
2(1) – 3 = A(1 + 1)(2 + 3)
⇒ – 1 = A(2)(5) ⇒ A = – $\frac{1}{10}$.
Putting x = – 1 in (1), we get
– 2 – 3 = B(- 1 – 1)(- 2 + 3)
⇒ – 5 = B(- 2)(1) ⇒ B = $\frac{5}{2}$
Putting x = – $\frac{3}{2}$ in (1), we get

Question 11.
$\frac{5 x}{(x+1)\left(x^{2}-4\right)}$
Solution:
Let $\frac{5 x}{(x+1)\left(x^{2}-4\right)}$ = $\frac{5x}{(x+1)(x+2)(x-2)}$
= $\frac{A}{x+1}$ + $\frac{B}{x+2}$ + $\frac{C}{x-2}$
⇒ 5x = A(x + 2)(x – 2) + B(x + 1)(x – 2) + C(x + 1)(x + 2) ………….. (1)
Putting x = – 1 in (1), we get
– 5 = A(- 1 + 2)(- 1 – 2) ⇒ A = $\frac{5}{3}$.
Putting x = – 2 in (1), we get
– 10 = B(- 2 + 1)(- 2 – 2) ⇒ B = – $\frac{5}{2}$
Putting x = 2 in (1), we get
10 = C(2 + 1)(2 + 2) ⇒ C = $\frac{5}{6}$.

Question 12.
$\frac{x^{3}+x+1}{x^{2}-1}$
Solution:
Since $\frac{x^{3}+x+1}{x^{2}-1}$ is an improper fraction, therefore we convert it into a proper fraction.
Dividing the polynomial function x³ + x + 1 by x² – 1 by long division, we get

Putting x = – 1 in (2), we get
⇒ 2x + 1 = A(- 1 – 1) ⇒ A = $\frac{-1}{-2}$ = $\frac{1}{2}$.
Putting x = 1 in (2), we get
2 + 1 = B(1 + 1) ⇒ B = $\frac{3}{2}$.
∴ $\frac{2 x+1}{x^{2}-1}$ = $\frac{1}{2(x+1)}$ + $\frac{3}{2(x-1)}$ …………. (3)
From (1) and (3),

Question 13.
$\frac{2}{(1-x)\left(1+x^{2}\right)}$
Solution:
Let $\frac{2 x+1}{x^{2}-1}$ = $\frac{A}{1-x}$ + $\frac{B x+C}{1+x^{2}}$.
⇒ 2 = A(1 + x²) + (Bx + C)(1 – x) …………… (1)
Putting x = 1 in (1), we get:
2 = A(1 + 1) ⇒ A = 1.
Comparing co-efficients of x² and the constant terms on both sides, we get:
0 = A – B and 2 = A + C
⇒ B = A = 1 and 2 = 1 + C.
⇒ C = 1.

Question 14.
$\frac{3 x-1}{(x+2)^{2}}$
Solution:

Question 15.
$\frac{1}{x^{4}-1}$
Solution:

Putting x = – 1 in (1) we get:
1 = A(- 1 – 1)(1 + 1)
⇒ 1 = A(- 4) ⇒ A = – $\frac{1}{4}$
Putting x = 1 in (1), we get:
1 = B(1 + 1)(1 + 1)
⇒ 1 = B(2)(2) ⇒ B = $\frac{1}{4}$.
Comparing the coefficients of x³ and constants in (1) on both sides, we get
0 = A + B + C

Question 16.
$\frac{1}{x\left(x^{n}+1\right)}$
Solution:
⇒ 1 = A(t + 1) + Bt …………….. (2)
Putting t = 0 in (2), we get:
1 = B(- 1) ⇒ B = – 1
∴ (1) gives:

Question 17.
$\frac{cosx}{(1-sin2x)(2-sinx)}$
Solution:
Put sin x = t so that cos x dx = dt.

⇒ 1 = A(2 – t) + B(1 – t) …………….. (2)
Putting t = 1 in (2) we get:
1 = A(2 – 1) ⇒ A = 1.
Putting t = 2 in (2), we get:
1 = B(1 – 2) ⇒ B = – 1.
∴ (1) gives:

Question 18.
$\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$
Solution:

Consider partial fractions of
$\frac{(2y+5)}{(y+3)(y+5)}$ = $\frac{A}{y+3}$ + $\frac{B}{y+4}$
Put y = – 3 in (2), – 6 + 5 = A.1 ∴ A = – 1.
Put y = – 4 in (2), – 8 + 5 = B(- 4 + 3)
or – 3 = – B ∴ B = 3.
Putting these values in (1), we get

Integrated both sides

Question 19.
$\frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}$
Solution:

Put t = – 1, 1 = A(- 1 + 3) = 2A ∴ A = $\frac{1}{2}$
Put t = – 3, 1 = B(- 3 + 1) = – 2B ∴ B = – $\frac{1}{2}$.
∴ $\frac{1}{(t+1)(t+3)}$ = $\frac{1}{2}$
Integrating both sides

Question 20.
$\frac{1}{x\left(x^{4}-1\right)}$
Solution:

Question 21.
$\frac{1}{e^{x}-1}$
Solution:

Choose the correct answer in each of the following questions 22 and 23:
Question 22.
∫ $\frac{xdx}{(x-1)(x-2)}$ equals
(A) log |$\frac{(x-1)^{2}}{x-2}$| + C
(B) log |$\frac{(x-2)^{2}}{x-1}$| + C
(C) log |($\frac{(x-1)^{2}}{x-2}$| + 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)
Put x = 1, 1 = A(1 – 2) = – A ∴ A = – 1
Put x = 2, 2 = B(2 – 1) = B ∴ B = 2.

∴ Part (B) is the correct answer.

Question 23.
∫ $\frac{d x}{x\left(x^{2}+1\right)}$ equals
(A) log|x| – $\frac{1}{2}$ log |x² + 1| + C
(B) log |x| + $\frac{1}{2}$ log|x² + 1| + C
(C) – log |x| + $\frac{1}{2}$ log |x² + 1| + C
(D) $\frac{1}{2}$ log|x| + log |x² + 1| + C
Solution:

∴ Part (A) is the correct answer.