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

\(\frac{x}{(x+1)(x+2)}\) = \(\frac{A}{x+1}\) + \(\frac{B}{x+2}\)

âˆ´ x = A(x + 2) + B(x + 1)

Put x = – 1, – 1 = A(- 1 + 2) + 0 = A âˆ´ A = – 1.

Put x = – 2, – 2 = 0 + B(- 2 + 1) = – B âˆ´ B = 2.

âˆ´ \(\frac{x}{(x+1)(x+2)}\) = – \(\frac{1}{x+1}\) + \(\frac{2}{x+2}\).

Integrating both sides, we get

Question 2.

\(\frac{1}{x^{2}-9}\)

Solution:

Let \(\frac{1}{x^{2}-9}\) = \(\frac{1}{(x-3)(x+3)}\) = \(\frac{A}{x-3}\) + \(\frac{B}{x+3}\).

â‡’ 1 = A(x + 3) + B(x – 3) ……………… (1)

Putting x = 3 in (1), we get:

1 = A(3 + 3) â‡’ A = \(\frac{1}{6}\).

Putting x = – 3 in (1), we get:

1 = B(- 3 – 3) â‡’ B = – \(\frac{1}{6}\)

Question 3.

\(\frac{3x-1}{(x-1)(x-2)(x-3)}\)

Solution:

Let \(\frac{3x-1}{(x-1)(x-2)(x-3)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{x-2}\) + \(\frac{C}{x-3}\).

â‡’ 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)

â‡’ 2 = A(- 1)(- 2) â‡’ A = 1.

Putting x = 2 in (1), we get:

6 – 1 = B(2 – 1)(2 – 3)

â‡’ 5 = B(1)(- 1) â‡’ B = – 5.

Putting x = 3 in (1), we get:

9 – 1 = C(3 – 1)(3 – 2)

â‡’ 8 = C(2)(1) â‡’ 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 \(\frac{x}{(x-1)(x-2)(x-3)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{x-2}\) + \(\frac{C}{x-3}\).

â‡’ 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)

â‡’ A = \(\frac{1}{2}\)

Putting x = 2 in (1), we get:

2 = B(2 – 1)(2 – 3) â‡’ B = – 2

Putting x = 3 in (1), we get:

3 = C(3 – 1)(3 – 2) â‡’ 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 – x^{2} by x – 2x^{2} by long division method, we get

â‡’ x – 2 = A(2x – 1) + Bx ………………. (2)

Puttting x = 0 in (2), we get:

– 2 = A(- 1) â‡’ 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

\(\frac{x}{\left(x^{2}+1\right)(x-1)}\) = \(\frac{A}{x-1}\) + \(\frac{B x+C}{x^{2}-1}\).

âˆ´ x = A(x^{2} + 1) + (Bx + C)(x – 1)

= A(x^{2} + 1) + B(x^{2} – x) + C(x – 1)

Put x = 1, 1 = A(1 + 1) â‡’ 1 = 2A or A = \(\frac{1}{2}\)

Comparing the coefficients of x^{2}, we get

0 = A + B

âˆ´ B = – A = – \(\frac{1}{2}\).

Comparing the constants, we get

0 = A – C âˆ´ C = A = \(\frac{1}{2}\).

Question 8.

\(\frac{x}{(x-1)^{2}(x+2)}\)

Solution:

Let \(\frac{x}{(x-1)^{2}(x+2)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{(x-1)^{2}}\) + \(\frac{C}{x+2}\).

â‡’ 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) â‡’ B = \(\frac{1}{3}\)

Putting x = – 2 in (1), we get:

– 2 = C(- 2 – 1)^{2} â‡’ C = – \(\frac{-2}{9}\)

Comparing co-efficients of x^{2} on both sides of (1), we have:

0 = A + C â‡’ 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 \(\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}\) = \(\frac{2x-3}{(x-1)(x+1)(2x+3)}\)

= \(\frac{A}{x-1}\) + \(\frac{B}{x+1}\) + \(\frac{C}{2x+3}\)

â‡’ 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)

â‡’ – 1 = A(2)(5) â‡’ A = – \(\frac{1}{10}\).

Putting x = – 1 in (1), we get

– 2 – 3 = B(- 1 – 1)(- 2 + 3)

â‡’ – 5 = B(- 2)(1) â‡’ 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 \(\frac{5 x}{(x+1)\left(x^{2}-4\right)}\) = \(\frac{5x}{(x+1)(x+2)(x-2)}\)

= \(\frac{A}{x+1}\) + \(\frac{B}{x+2}\) + \(\frac{C}{x-2}\)

â‡’ 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) â‡’ A = \(\frac{5}{3}\).

Putting x = – 2 in (1), we get

– 10 = B(- 2 + 1)(- 2 – 2) â‡’ B = – \(\frac{5}{2}\)

Putting x = 2 in (1), we get

10 = C(2 + 1)(2 + 2) â‡’ 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 x^{3} + x + 1 by x^{2} – 1 by long division, we get

Putting x = – 1 in (2), we get

â‡’ 2x + 1 = A(- 1 – 1) â‡’ A = \(\frac{-1}{-2}\) = \(\frac{1}{2}\).

Putting x = 1 in (2), we get

2 + 1 = B(1 + 1) â‡’ B = \(\frac{3}{2}\).

âˆ´ \(\frac{2 x+1}{x^{2}-1}\) = \(\frac{1}{2(x+1)}\) + \(\frac{3}{2(x-1)}\) …………. (3)

From (1) and (3),

Question 13.

\(\frac{2}{(1-x)\left(1+x^{2}\right)}\)

Solution:

Let \(\frac{2 x+1}{x^{2}-1}\) = \(\frac{A}{1-x}\) + \(\frac{B x+C}{1+x^{2}}\).

â‡’ 2 = A(1 + x^{2}) + (Bx + C)(1 – x) …………… (1)

Putting x = 1 in (1), we get:

2 = A(1 + 1) â‡’ A = 1.

Comparing co-efficients of x^{2} and the constant terms on both sides, we get:

0 = A – B and 2 = A + C

â‡’ B = A = 1 and 2 = 1 + C.

â‡’ 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)

â‡’ 1 = A(- 4) â‡’ A = – \(\frac{1}{4}\)

Putting x = 1 in (1), we get:

1 = B(1 + 1)(1 + 1)

â‡’ 1 = B(2)(2) â‡’ B = \(\frac{1}{4}\).

Comparing the coefficients of x^{3} and constants in (1) on both sides, we get

0 = A + B + C

Question 16.

\(\frac{1}{x\left(x^{n}+1\right)}\)

Solution:

â‡’ 1 = A(t + 1) + Bt …………….. (2)

Putting t = 0 in (2), we get:

1 = B(- 1) â‡’ B = – 1

âˆ´ (1) gives:

Question 17.

\(\frac{cosx}{(1-sin2x)(2-sinx)}\)

Solution:

Put sin x = t so that cos x dx = dt.

â‡’ 1 = A(2 – t) + B(1 – t) …………….. (2)

Putting t = 1 in (2) we get:

1 = A(2 – 1) â‡’ A = 1.

Putting t = 2 in (2), we get:

1 = B(1 – 2) â‡’ B = – 1.

âˆ´ (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 âˆ´ A = – 1.

Put y = – 4 in (2), – 8 + 5 = B(- 4 + 3)

or – 3 = – B âˆ´ 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 âˆ´ A = \(\frac{1}{2}\)

Put t = – 3, 1 = B(- 3 + 1) = – 2B âˆ´ B = – \(\frac{1}{2}\).

âˆ´ \(\frac{1}{(t+1)(t+3)}\) = \(\frac{1}{2}\)

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.

âˆ« \(\frac{xdx}{(x-1)(x-2)}\) 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 \(\frac{x}{(x-1)(x-2)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{x-2}\)

âˆ´ x = A(x – 2) + B(x – 1)

Put x = 1, 1 = A(1 – 2) = – A âˆ´ A = – 1

Put x = 2, 2 = B(2 – 1) = B âˆ´ B = 2.

âˆ´ Part (B) is the correct answer.

Question 23.

âˆ« \(\frac{d x}{x\left(x^{2}+1\right)}\) 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:

âˆ´ Part (A) is the correct answer.