Ncert Solutions For Class 12 Maths Ex 7.5

Ncert Solutions For Class 12 Maths Chapter 7 Ex 7.5

Question 1:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u}{(u + 1) (u + 2)}$$

Suppose, $$\frac{u}{(u + 1) (u + 2)} = \frac{A}{u + 1} + \frac{B}{u + 2} \\ => u = A (u + 2) + B (u + 1)$$

Equate the coefficients of u and the constants on both the sides, we get,

A + B = 1

2 A + B = 0

On solving, we get,

A = – 1 and B = 2

$$\frac{u}{(u + 1) (u + 2)} = \frac{- 1}{u + 1} + \frac{2}{u + 2} \\ => \int \frac{u}{(u + 1) (u + 2)} \;du = \frac{- 1}{u + 1} + \frac{2}{u + 2} \;du \\ = – log \left | u + 1 \right | + 2 log \left | u + 2 \right | + C \\ = log \left ( u + 2 \right ) ^{2} – log \left | u + 1 \right | + C \\ = log \frac{(u + 1) ^{2}}{(u + 1)} + C$$

Question 2:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{u ^{2} – 9}$$

Suppose, $$\frac{1}{(u + 3) (u – 3)} = \frac{A}{u + 3} + \frac{B}{u – 3} \\ 1 = A (u – 3) + B (u + 3)$$

Equate the coefficients of u and the constants on both the sides, we get,

A + B = 0

– 3 A + 3 B = 1

On solving, we get

$$A = – \frac{1}{6} \; and\; B = \frac{1}{6} \\ \frac{1}{(u + 3) (u – 3)} = \frac{- 1}{6 (u + 3)} + \frac{1}{6 (u – 3)} \\ => \int \frac{1}{u ^{2} – 9} \;du = \int \left ( \frac{- 1}{6 (u + 3)} + \frac{1}{6 (u – 3)} \right ) \;du \\ = – \frac{1}{6} log \left | u + 3 \right | + \frac{1}{6} log \left | u – 3 \right | + C \\ = \frac{1}{6} log \left | \frac{(u – 3)}{(u + 3)} \right | + C$$

Question 3:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)}$$

Suppose, $$\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} = \frac{A}{(u – 1)} + \frac{B}{(u – 2} + \frac{C}{(u – 3)} \\ 3 u – 1 = A (u – 2) (u – 3) + B (u – 1) (u – 3) + C (u – 1) (u – 2) ….. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– 5 A – 4 B – 3 C = 1

6 A + 3 B + 2 C = – 1

On solving, we get,

A = 1, B = – 5, and C = 4

$$\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} = \frac{1}{(u – 1)} – \frac{5}{(u – 2} + \frac{4}{(u – 3)} \\ \int \frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} \;du = \int \left \{ \frac{1}{(u – 1)} – \frac{5}{(u – 2} + \frac{4}{(u – 3)} \right \} \;du \\ = log \left | u – 1 \right | – 5 log \left | u – 2 \right | + 4 log \left | u – 3 \right | + C$$

Question 4:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u}{(u – 1) (u – 2) (u – 3)}$$

Suppose, $$\frac{u}{(u – 1) (u – 2) (u – 3)} = \frac{A}{(u – 1)} + \frac{B}{(u – 2} + \frac{C}{(u – 3)} \\ u = A (u – 2) (u – 3) + B (u – 1) (u – 3) + C (u – 1) (u – 2) ….. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– 5 A – 4 B – 3 C = 1

6 A + 3 B + 2 C = 0

On solving, we get,

$$A = \frac{1}{2}, \;B = – 2 \;and \;C = \frac{3}{2} \\ \frac{u}{(u – 1) (u – 2) (u – 3)} = \frac{1}{2 (u – 1)} – \frac{2}{(u – 2} + \frac{3}{2 (u – 3)} \\ \int \frac{u}{(u – 1) (u – 2) (u – 3)} \;du = \int \left \{ \frac{1}{2 \;(u – 1)} – \frac{2}{(u – 2)} + \frac{3}{2 (u – 3)} \right \} \;du \\ = \frac{1}{2}\; log \left | u – 1 \right | – 2 log \left | u – 2 \right | + \frac{3}{2} log \left | u – 3 \right | + C$$

Question 5:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{2 u}{u ^{2} + 3 u + 2}$$

Suppose, $$\frac{2 u}{u ^{2} + 3 u + 2} = \frac{A}{(u + 1)} + \frac{B}{(u + 2)} \\ 2 u = A (u + 2) + B (u + 1) …. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 2

2 A + B = 0

On solving, we get,

A = – 2 and B = 4

$$\frac{2 u}{(u + 1) (u + 2)} = \frac{- 2}{(u + 1)} + \frac{4}{(u + 1)} \\ \int \frac{2 u}{(u + 1) (u + 2)} \;du = \int \left \{ \frac{4}{(u + 1)} – \frac{2}{(u + 1)}\right \} \;du \\ = 4 log \left | u + 2 \right | – 2 log \left | u + 1 \right | + C$$

Question 6:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1 – u ^{2}}{u (1 – 2 u)}$$

$$\frac{1 – u ^{2}}{u (1 – 2 u)}$$ is not a proper fraction.

Dividing (1 – u 2) by u (1 – 2 u), we get,

$$\frac{1 – u ^{2}}{u (1 – 2 u)} = \frac{1}{2} + \frac{1}{2} \left ( \frac{2 – u}{u (1 – 2 u)} \right ) \\ Suppose,\; \frac{2 – u}{u (1 – 2 u)} = \frac{A}{u} + \frac{B}{(1 -2 u)} \\ (2 – u) = A (1 – 2 u) + B u ……. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

– 2 A + B = – 1

And A = 2

On solving, we get,

A = 2 and B = 3

$$\frac{2 – u}{u (1 – 2 u)} = \frac{2}{u} + \frac{3}{(1 -2 u)} \\ Using\; in\; equation\; (1), we\; get, \\ \frac{1 – u ^{2}}{u (1 – 2 u)} = \frac{1}{2} + \frac{1}{2} \left \{ \frac{2}{u} + \frac{3}{(1 -2 u)} \right \} \\ = \frac{u}{2} + log \left | u \right | + \frac{3}{2 (- 2)}\; log \left | 1 – 2u \right | + C \\ = \frac{u}{2} + log \left | u \right | – \frac{3}{4}\; log \left | 1 – 2u \right | + C$$

Question 7:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u}{(u ^{2} + 1) (u – 1)}$$

Suppose, $$\frac{u}{(u ^{2} + 1) (u – 1)} = \frac{A u + B}{(u ^{2} + 1)} + \frac{C}{u – 1} \\ u = (A u + B) (u – 1) + C (u^{2} + 1) \\ u = A u ^{2} – A u + B u – B + C u^{2} + C$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + C = 0

– A + B = 1

– B + C = 0

On solving, we get,

$$A = – \frac{1}{2},\; B = \frac{1}{2}, and\; C = \frac{1}{2} \\ Using\; equation\; (1),\; we\; get\; \\ \frac{u}{(u ^{2} + 1) (u – 1)} = \frac{\left ( – \frac{1}{2} u + \frac{1}{2} \right )}{(u ^{2} + 1)} + \frac{\frac{1}{2}}{(u – 1)} \\$$ $$\int \frac{u}{(u ^{2} + 1) (u – 1)} \;du = – \frac{1}{2} \int \frac{u}{(u ^{2} + 1)} \;du + \frac{1}{2} \int \frac{1}{(u ^{2} + 1)} + \frac{1}{2} \int \frac{1}{2} \int \frac{1}{u – 1} \;du \\ = – \frac{1}{4} \int \frac{2 u}{(u ^{2} + 1)} + \frac{1}{2} tan ^{- 1} u + \frac{1}{2} log \left | u – 1 \right | + C \\ \int \frac{2 u}{(u ^{2} + 1)} \;du, let\; (u ^{2} + 1) = z => 2u\; du = dz \\ \int \frac{2 u}{(u ^{2} + 1)} \;du = \int \frac{dz}{z} = log \left | z \right | = log\; \left | u^{2} + 1 \right | \\ \int \frac{u}{(u ^{2} + 1) (u – 1)} \;du = – \frac{1}{4} log \left | (u ^{2} + 1) \right | + \frac{1}{2} tan ^{- 1}u + \frac{1}{2} log \left | u – 1 \right | + C \\ = \frac{1}{2} log \left | u – 1 \right | – \frac{1}{4} log \left | (u ^{2} + 1) \right | + \frac{1}{2} tan ^{- 1}u + C$$

Question 8:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u}{(u – 1)^{2} (u + 2)}$$

$$\frac{u}{(u – 1) c^{2} (u + 2)} = \frac{A}{(u – 1)} + \frac{B}{(u – 1) ^{2}} + \frac{C}{u + 2} \\ u = A (u – 1) (u + 2) + B (u + 2) + C (u – 1)^{2} \\$$

Putting u = 1, we get,

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + C = 0

A + B – 2 C = 1

– 2 A + 2 B + C = 0

On solving, we get,

$$A = \frac{2}{9},\; B = \frac{1}{3} \;and\; C = – \frac{2}{9}$$ $$\frac{u}{(u – 1)^{2} (u + 2)} = \frac{2}{9 \;(u – 1)} + \frac{1}{3 \;(u – 1) ^{2}} – \frac{2}{9 (u + 2)} \\ \int \frac{u}{(u – 1)^{2} (u + 2)} \;du = \frac{2}{9} \int \frac{1}{(u – 1)} \;du + \frac{1}{3} \int \frac{1}{(u – 1) ^{2}} \;du – \frac{2}{9} \int \frac{1}{(u + 2)} \;du \\ = \frac{2}{9} log\; \left | u – 1 \right | + \frac{1}{3} \left ( \frac{- 1}{u – 1} \right ) – \frac{2}{9} log \left | u + 2 \right | + C \\ = \frac{2}{9} log\; \left | \frac{u – 1}{u + 2} \right | – \frac{1}{3 (u – 1)} + C$$

Question 9:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{3 u + 5}{u^{3} – u^{2} – u + 1}$$

$$\frac{3 u + 5}{u^{3} – u^{2} – u + 1} = \frac{3 u + 5}{(u – 1)^{2} (u + 1)} \\ Suppose,\; \frac{3 u + 5}{(u – 1)^{2} (u + 1)} = \frac{A}{(u – 1)} + \frac{B}{(u – 1)^{2}} + \frac{C}{(u + 1)} \\ 3 u + 5 = A (u – 1) (u + 1) + B (u + 1) + C (u – 1) ^{2} \\ 3 u + 5 = A (u^{2} – 1) + B (u + 1) + C (u^{2} + 1 – 2 u) \\$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + C = 0

B – 2 C = 3

– A + B + C = 5

On solving, we get,

B = 4

$$A = – \frac{1}{2} \;and\; C = \frac{1}{2} \\ \frac{3 u + 5}{(u – 1)^{2} (u + 1)} = \frac{- 1}{2 (u – 1)} + \frac{4}{(u – 1)^{2}} + \frac{1}{2 (u + 1)} \\ \int \frac{3 u + 5}{(u – 1)^{2} (u + 1)} \;du = – \frac{1}{2} \int \frac{1}{(u – 1)} \;du + 4 \int \frac{1}{(u – 1)^{2}} \;du + \frac{1}{2} \int \frac{1}{(u + 1)} \;du \\ = – \frac{1}{2} log\; \left | u – 1 \right | + 4 \left ( \frac{- 1}{u – 1} \right ) + \frac{1}{2} log \left | u + 1 \right | + C \\ = \frac{1}{2} log \left | \frac{u + 1}{u – 1} \right | – \frac{4}{(u – 1)} + C$$

Question 10:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{2 u – 3}{(u^{2} – 1) (2u + 3)}$$

$$\frac{2 u – 3}{(u^{2} – 1) (2u + 3)} = \frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)}$$

Suppose, $$\frac{2 u – 3}{(u^{2} – 1) (2u + 3)} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} + \frac{C}{(2u + 3)} \\ (2u – 3) = A\; (u – 1)(2 u + 3) + B\; (u + 1) (2 u + 3) + C\; (u + 1) (u – 1) \\ (2u – 3) = A\; (2 u^{2} + u – 3) + B\; (2 u^{2} + 5u + 3) + C\; (u^{2} – 1) \\ (2u – 3) = (2A + 2B + C) u^{2} + (A + 5B)u + (- 3A + 3B – C)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

2A + 2B + C = 1

A + 5B = 2

– 3A + 3B – C = – 3

On solving, we get,

$$\frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)} = \frac{5}{2 (u + 1)} – \frac{1}{10 (u – 1)} – \frac{24}{5 (2u + 3)} \\ \frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)} = \frac{5}{2} \int \frac{1}{(u + 1)} \;du – \frac{1}{10} \int \frac{1}{ (u – 1)} \;du – \frac{24}{5} \int \frac{1}{ (2u + 3)} \;du \\ = \frac{5}{2} log\; \left | u + 1 \right | – \frac{1}{10} log\; \left | u – 1 \right | – \frac{24}{5 \times 2} log\; \left | 2u + 3 \right | + C \\ = \frac{5}{2} log\; \left | u + 1 \right | – \frac{1}{10} log\; \left | u – 1 \right | – \frac{12}{5} log\; \left | 2u + 3 \right | + C$$

Question 11:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{5 u}{(u + 1) (u^{2} – 4)}$$

$$\frac{5 u}{(u + 1) (u^{2} – 4)} = \frac{5 u}{(u + 1)(u + 2)(u – 2)} \\ Suppose,\; \frac{5 u}{(u + 1)(u + 2)(u – 2)} = \frac{A}{(u + 1)} + \frac{B}{(u + 2)} + \frac{C}{(u – 2)} \\ 5u = A (u + 2)(u – 2) + B (u + 1)(u – 2) + C (u + 1)(u + 2) …. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– B + 3C = 5 and

– 4A – 2B + 2C = 0

On solving, we get,

$$A = \frac{5}{3}, B = – \frac{5}{2} \;and\; C = \frac{5}{6} \\ \frac{5 u}{(u + 1)(u + 2)(u – 2)} = \frac{5}{3 (u + 1)} – \frac{5}{2 (u + 2)} + \frac{5}{6 (u – 2)} \\ \int \frac{5 u}{(u + 1)(u + 2)(u – 2)} \;du = \frac{5}{3} \frac{1}{(u + 1)} \;du – \frac{5}{2} \frac{1}{(u + 2)} \;du + \frac{5}{6} \frac{1}{(u – 2)} \;du \\ = \frac{5}{3} log\; \left | u + 1 \right | – \frac{5}{2} log\; \left | u + 2 \right | + \frac{5}{6} log\; \left | u – 2 \right | + C$$

Question 12:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u^{3} + u + 1}{u^{2} – 1}$$

$$\frac{u^{3} + u + 1}{u^{2} – 1}$$ is not a proper fraction.

So, dividing (u3 + u + 1) by u2 – 1, we get,

$$\frac{u^{2} + u + 1}{u^{2} – 1} = u + \frac{2u + 1}{u^{2} – 2} \\ Suppose,\; \frac{2u + 1}{u^{2} – 2} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} \\ 2u + 1 = A (u – 1) + B (u + 1) ….. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 2

– A + B = 1

On solving, we get,

$$A = \frac{1}{2} \;and\; B = \frac{3}{2} \\ \frac{u^{2} + u + 1}{u^{2} – 1} = u + \frac{1}{2 (u + 1)} + \frac{3}{2 (u – 1)} \\ Integrating\; on\; both\; the\; sides,\; we\; get, \int \frac{u^{2} + u + 1}{u^{2} – 1} \;du = \int u \;du + \frac{1}{2} \int \frac{1}{u + 1} \;du + \frac{3}{2} \frac{1}{(u – 1)} \;du \\ = \frac{u ^{2}}{2} + \frac{1}{2} log\; \left | u + 1 \right | – \frac{3}{2} log\; \left | u – 1 \right | + C$$

Question 13:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{2}{(1 – u)(1 + u ^{2})}$$

Suppose, $$\frac{2}{(1 – u)(1 + u ^{2})} = \frac{A}{1 – u} + \frac{Bu + C}{1 + u ^{2}} \\ 2 = A (1 + u ^{2}) + (Bu + C)(1 – u) \\ 2 = A + A\; u^{2} + Bu – B\; u^{2} + C – Cu$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A – B = 0

B – C = 0

A + C = 2

On solving, we get,

A = 1, B = 1 and C = 1

$$\frac{2}{(1 – u)(1 + u ^{2})} = \frac{1}{1 – u} + \frac{u + 1}{1 + u ^{2}} \\ \int \frac{2}{(1 – u)(1 + u ^{2})} \;du = \int \frac{1}{1 – u} \;du + \int \int \frac{u}{1 + u ^{2}} \;du + \int \frac{1}{1 + u ^{2}} \;du \\ = – log\; \left | u – 1 \right | + \frac{1}{2} log\; \left | 1 + u^{2} \right | + tan ^{- 1} \;u + C$$

Question 14:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{3u – 1}{(u + 2)^{2}}$$

Suppose, $$\frac{3u – 1}{(u + 2)^{2}} = \frac{A}{(u + 2)} + \frac{B}{(u + 2)^{2}} \\ 3u – 1 = A (u + 2) + B$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = 3

2A + B = – 1

B = – 7

$$\frac{3u – 1}{(u + 2)^{2}} = \frac{3}{(u + 2)} – \frac{7}{(u + 2)^{2}} \\ 3u – 1 = A (u + 2) + B \frac{3u – 1}{(u + 2)^{2}} = 3 \int \frac{1}{(u + 2)} \;du – 7 \int \frac{u}{(u + 2)^{2}} \;du \\ = 3\; log\; \left | u + 2 \right | – 7 \left ( \frac{- 1}{(u\; + \;2)} \right ) + C \\ = 3\; log\; \left | u + 2 \right | + \left ( \frac{7}{(u\; + \;2)} \right ) + C$$

Question 15:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{u^{4} – 1}$$

$$\frac{1}{u^{4} – 1} = \frac{1}{(u^{2} – 1)(u^{4} + 1)} = \frac{1}{(u + 1) (u – 1)(1 + u^{2})} \\ Suppose,\; \frac{1}{(u + 1) (u – 1)(1 + u^{2})} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} + \frac{Cu + D}{(u^{2} + 1)} \\ 1 = A (u + 1)(u^{2} + 1) + B (u + 1)(u^{2} + 1) + (Cu + D) (u^{2} – 1) \\ 1 = A (u^{3} + u – u^{2} – 1) + B (u^{3} + u + u^{2} + 1) + Cu^{2} + Du^{2} – Cu – D \\ 1 = (A + B + C) u^{3} + (- A + B + D) u^{2} + (A + B – C)u + (- A + B – D)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– A + B + D = 0

A + B – C = 0

– A + B – D = 1

On solving, we get,

$$A = – \frac{1}{4}, B = \frac{1}{4}, C = 0 \;and\; D = – \frac{1}{2} \\ \frac{1}{u^{4} – 1} = \frac{- 1}{4 (u + 1)} + \frac{1}{4 (u – 1)} – \frac{1}{2 (u^{2} + 1)} \\ \int \frac{1}{u^{4} – 1} \;du = – \frac{1}{4} \int \frac{1}{(u + 1)} \;du + \frac{1}{4} \int \frac{1}{(u – 1)} \;du + \frac{1}{2} \int \frac{1}{(u^{2} + 1)} \\ = – \frac{1}{4} log\; \left | u + 1 \right | + \frac{1}{4} log\; \left | u – 1 \right | – \frac{1}{2} tan^{- 1} \;u + C \\ = \frac{1}{4} log\; \left | \frac{ u – 1}{u + 1} \right | – \frac{1}{2} tan^{- 1} \;u + C$$

Question 16:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{u (u^{m} + 1)}$$

[Hint: multiply denominator and numerator by u n – 1 and put un = z]

$$\frac{1}{u (u^{m} + 1)}$$

Multiplying denominator and numerator by u n – 1, we get,

$$\frac{1}{u (u^{m} + 1)} = \frac{u ^{m – 1}}{u ^{m – 1}. u (u^{m} + 1)} = \frac{u ^{m – 1}}{u ^{m} (u^{m} + 1)} \\ Suppose,\; u^{m} = z => u ^{m – 1} \;du = dz \\ \int \frac{1}{u (u^{m} + 1)} \;du = \int \frac{u ^{m – 1}}{u ^{m} (u^{m} + 1)} \;du = \frac{1}{m} \int \frac{1}{z (z + z)} \;du \\ Suppose,\; \frac{1}{z (z + z)} = \frac{A}{z} + \frac{B}{(z + 1)} \\ 1 = A (1 + z) + Bz \\$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = 1 and B = – 1

$$\frac{1}{z (z + z)} = \frac{1}{z} – \frac{1}{(z + 1)} \\ \int \frac{1}{u (u^{m} + 1)} \;du = \frac{1}{m} \int \left \{ \frac{1}{z} – \frac{1}{(z + 1)} \right \} + C \\ = \frac{1}{m} \left [ log\; \left | u^{m} \right | – log\; \left | u^{m} + 1 \right | \right ] + C \\ = \frac{1}{m} log\; \left | \frac{u^{m}}{u^{m} + 1} \right |$$

Question 17:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)}$$

[Hint: Put sin u = z]

$$\frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)}$$

Suppose, sin u = z => cos u du = dz

$$\int \frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)} \;du = \int \frac{dz}{(1 – z)(2 – z)} \\ Suppose,\; \frac{1}{(1 – z)(2 – z)} = \frac{A}{(1 – z)} + \frac{B}{(2 – z)} \\ 1 = A (2 – z) + B (1 – z)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

– 2A – B = 0

2A + B = 1

On solving, we get,

A = 1 and B = – 1

$$\frac{1}{(1 – z)(2 – z)} = \frac{1}{(1 – z)} – \frac{1}{(2 – z)} \\ \int \frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)} \;du = \int \left \{ \frac{1}{(1 – z)} – \frac{1}{(2 – z)} \right \} \;dz \\ = – log\; \left | 1 – z \right | + log\; \left | 2 – z \right | + C \\ = log\; \left | \frac{2 – z}{1 – z} \right | + C \\ = log\; \left | \frac{2 – sin\; u}{1 – sin\; u} \right | + C$$

Question 18:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)}$$

$$\frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} = 1 – \frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} \\ Suppose,\; \frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} = \frac{Au + B}{(u^{2} + 3)} + \frac{Cu + D}{(u^{2} + 4)} \\ (4 u^{2} + 10) = (Au + B)(u^{2} + 4) + (Cu + D)(u^{2} + 3) \\ (4 u^{2} + 10) = A u^{3} + 4 Au + B u^{2} + 4 B + C u^{3} + 3 Cu + D u^{2} + 3D \\ (4 u^{2} + 10) = (A + C) u^{3} + (B + D) u^{2} + (4A + 3C) u + (4B + 3D)$$

Equate the coefficients of u3, u2, u and the constants on both the sides, we get,

A + C = 0

B + D = 4

4 A + 3 C = 0

4 B + 3 D = 10

On solving, we get,

A = 0, B = – 2, C = 0 and D = 6

$$\frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} = \frac{- 2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \\ \frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} = 1 – \left ( \frac{- 2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \right ) \\ \int \frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} \;du = \int \left \{ 1 + \frac{2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \right \} \\ = \int \left \{ 1 + \frac{2}{u ^{2} + (\sqrt{3})^{2}} – \frac{6}{u^{2} + 2^{2}} \right \} \;du \\ = u + 2 \left ( \frac{1}{\sqrt{3}} \;tan ^{- 1} \frac{u}{\sqrt{3}} \right ) – 6 \left ( \frac{1}{2} \;tan ^{- 1} \frac{u}{2} \right ) + C \\ = u + \frac{2}{\sqrt{3}} \;tan ^{- 1} \frac{u}{\sqrt{3}} – 3 \;tan ^{- 1} \frac{u}{2} + C$$

Question 19:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{2\; u}{(u^{2} + 1)(u^{2} + 3)}$$

$$\frac{2\; u}{(u^{2} + 1)(u^{2} + 3)}$$

Suppose, u2 = z

2u du = dz

$$\int \frac{2\; u}{(u^{2} + 1)(u^{2} + 3)} \;du = \int \frac{dz}{(z + 1)(z + 3)} …. (1) \\ Suppose,\; \frac{1}{(z + 1)(z + 3)} = \frac{A}{(z + 1)} + \frac{B}{(z + 3)} \\ 1 = A\; (z + 3) + B\; (z + 1) …. (1) \\$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 0

3 A + B = 1

On solving, we get,

$$A = \frac{1}{2} \;and\; B = – \frac{1}{2} \\ \frac{1}{(z + 1)(z + 3)} = \frac{1}{2\; (z + 1)} – \frac{1}{2\; (z + 3)} \\ \int \frac{2\; u}{(u^{2} + 1)(u^{2} + 3)} \;du = \int \left \{ \frac{1}{2\; (z + 1)} – \frac{1}{2\; (z + 3)} \right \} \;dz \\ = \frac{1}{2} log \left | (z + 1) \right | – \frac{1}{2} log \left | (z + 3) \right | + C \\ = \frac{1}{2} log \left | \frac{z + 1}{z + 3} \right | + C \\ = \frac{1}{2} log \left | \frac{u^{2} + 1}{u^{2} + 3} \right | + C$$

Question 20:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{u (u^{4} – 1)}$$

$$\frac{1}{u (u^{4} – 1)}$$

Multiplying denominator and numerator by u3, we get,

$$\frac{1}{u (u^{4} – 1)} = \frac{u^{3}}{u^{4} (u^{4} – 1)} \\ \int \frac{1}{u (u^{4} – 1)} \;du = \int \frac{u^{3}}{u^{4} (u^{4} – 1)} \;du \\ Suppose,\; u^{4} = t => 4 u^{3} \;du = dz \\ \int \frac{1}{u (u^{4} – 1)} \;du = \frac{1}{4} \int \frac{dz}{z (z – 1)} \\ Suppose,\; \frac{1}{z (z – 1)} = \frac{A}{z} + \frac{B}{z &##8211; 1} \\ 1 = A (z – 1) + Bz ….. (1) \\$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = – 1 and B = 1

$$\frac{1}{z (z – 1)} = \frac{- 1}{z} + \frac{1}{z – 1} \\ = \int \frac{1}{u (u^{4} – 1)} \;du = \frac{1}{4} \int \left ( \frac{- 1}{z} + \frac{1}{z – 1} \right ) \;dz \\ = \frac{1}{4} \left [ – log\; \left | z \right | + log\; \left | z – 1 \right | \right ] + C \\ = \frac{1}{4} log\; \left | \frac{z – 1}{z} \right | + C \\ = \frac{1}{4} log\; \left | \frac{u^{4} – 1}{u^{4}} \right | + C$$

Question 21:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{(e^{u} – 1)}$$

$$\frac{1}{(e^{u} – 1)}$$

Suppose, eu = z

eu du = dz

$$\int \frac{1}{(e^{u} – 1)} \;du = \int \frac{1}{z – 1} \times \frac{dz}{z} = \int \frac{1}{z (z – 1)} \;dz \\ Suppose,\; \frac{1}{z (z – 1)} = \frac{A}{z} + \frac{B}{z – 1} \\ 1 = A (z – 1) + Bz$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = – 1 and B = 1

$$\frac{1}{z (z – 1)} = \frac{- 1}{z} + \frac{1}{z – 1} \\ \int \frac{1}{z (z – 1)} \;du = log\; \left | \frac{z – 1}{z} \right | + C \\ = log\; \left | \frac{e^{u} – 1}{e^{u}} \right | + C$$

Question 22: Which of the following below is an integral of $$\frac{u \;du}{(u – 1)(u – 2)}$$

$$(a) log\; \left | \frac{(u – 1)^{2}}{u – 2} \right | + C \\ (b) log\; \left | \frac{(u – 2)^{2}}{u – 2} \right | + C \\ (c) log\; \left | (\frac{(u – 1)}{u – 2}) ^{2} \right | + C \\ (d) log\; \left | (u – 1)(u – 2) \right | + C$$

$$Suppose,\; \frac{u \;du}{(u – 1)(u – 2)} = \frac{A}{(u – 1)} + \frac{B}{u – 2} \\ u = A (u – 2) + B (u – 1) …. (1)$$

Equate the coefficients of u and the constants on both the sides, we get,

A = – 1 and B = 2

$$\frac{u \;du}{(u – 1)(u – 2)} = \frac{- 1}{(u – 1)} + \frac{2}{u – 2} \\ \int \frac{u \;du}{(u – 1)(u – 2)} \;d= \left \{ \frac{- 1}{(u – 1)} + \frac{2}{u – 2} \right \} \;du \\ = – log\; \left | u – 1 \right | + 2\; log\; \left | u – 2 \right | + C \\ = log\; \left | \frac{(u – 2) ^{2}}{u – 1} \right | + C$$

Hence, option (b) is the correct answer.

Question 23: Which of the following below is an integral of $$\int \frac{du}{u (u^{2} + 1)} \;du$$

$$(a) log\; \left | u \right | – \frac{1}{2} log\; (u^{2} + 1) + C \\ (b) log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\ (c) – log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\ (d) log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\$$

$$Suppose,\; \frac{1}{u (u^{2} + 1)} = \frac{A}{u} + \frac{Bu + C}{u^{2} + 1} \\ 1 = A (u^{2} + 1) + (Bu + C) u$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 0

C = 0

A = 1

On solving, we get,

A =1, B = – 1, and C = 0

$$\frac{1}{u (u^{2} + 1)} = \frac{1}{u} + \frac{- U}{u^{2} + 1} \\ \int \frac{1}{u (u^{2} + 1)} \;du = \int \left \{ \frac{1}{u} – \frac{u}{u^{2} + 1} \right \} \;du \\ = log\; \left | u \right | – \frac{1}{2} log\; \left | u^{2} + 1 \right | + C$$

Hence, option (a) is the correct answer.