Ncert Solutions For Class 12 Maths Ex 7.3

Ncert Solutions For Class 12 Maths Chapter 7 Ex 7.3

Question 1:

Obtain an integral (or anti – derivative) of the \(\frac{3 u^{2}}{u^{6} + 1}\)

Answer 1:

\(Suppose,\; u ^{3} = z \\ 3 u ^{2} du = dz \\ \int \frac{3 u^{2}}{u^{6} + 1} du = \int \frac{dz}{z ^{2} + 1} \\ = tan ^{- 1}\; z + C \\ = tan ^{- 1}\; u ^{3} + C\)

Question 2:

Obtain an integral (or anti – derivative) of the \(\frac{1}{\sqrt{1 + 4 u ^{2}}}\)

Answer 2:

\(Suppose,\; 2 u = z \\ 2\; du = dz \\ \int \frac{1}{\sqrt{1 + 4 u ^{2}}}\; du = \frac{1}{2} \int \frac{dz}{\sqrt{1 + z ^{2}}} \\ = \frac{1}{2} [log \left | z + \sqrt{1 + z ^{2}} \right |] + C \\ = \frac{1}{2} [log \left | 2 u + \sqrt{1 + 4 u^{2}} \right |] + C\)

Question 3:

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

Answer 3:

\(Suppose,\; 2 – u = z \\ – du = dz \\ \int \frac{1}{\sqrt{(2 – u)^{2} + 1}}\; du = – \int \frac{1}{\sqrt{z ^{2} + 1}} \;dz \\ = – [log \left | z + \sqrt{z ^{2} + 1} \right |] + C \\ = – [log \left | 2 – u + \sqrt{(2 – u) ^{2} + 1} \right |] + C \\ = log \left | \frac{1}{(2 – u) + \sqrt{u ^{2} – 4u + 5}} \right | + C\)

Question 4:

Obtain an integral (or anti – derivative) of the \(\frac{1}{\sqrt{9 – 25 u^{2}}}\)

Answer 4:

Suppose, 5u = z

5 du = dz

\(\int \frac{1}{\sqrt{9 – 25 u^{2}}}\; du = \frac{1}{5} \int \frac{1}{9 – z ^{2}}\; dz \\ = \frac{1}{5} \int \frac{1}{\sqrt{3 ^{2} – z ^{2}}}\; dz \\ = \frac{1}{5} sin ^{- 1} \;\frac{z}{3} + C \\ = \frac{1}{5} sin ^{- 1} \;\frac{5 u}{3} + C \\\)

Question 5:

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

Answer 5:

\(Suppose,\; \sqrt{2} u^{2} = z \\ 2 \sqrt{2}\; u\; du = dz \\ \int \frac{3 u}{1 + 2 u ^{4}}\; du = \frac{3}{2 \sqrt{2}} \int \frac{dz}{1 + z ^{2}} dz \\ = \frac{3}{2 \sqrt{2}} [tan ^{- 1} z] + C \\ = \frac{3}{2 \sqrt{2}} [tan ^{- 1} (\sqrt{2} u^{2})] + C\)

Question 6:

Obtain an integral (or anti – derivative) of the \(\frac{u ^{2}}{1 – u ^{6}}\)

Answer 6:

Suppose, u3 = z

3 u2 du = dz

\(\int \frac{u ^{2}}{1 – u ^{6}} du = \frac{1}{3} \int \frac{dz}{1 – z^{2}} \\ = \frac{1}{3} \left [ \frac{1}{2} log \left |\frac{1 + z}{1 – z} \right |\right ] + C \\ = \frac{1}{6} log \left | \frac{1 + u^{3}}{1 – u^{3}} \right | + C\)

Question 7:

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

Answer 7:

\(\int \frac{u – 1}{\sqrt{u ^{2} – 1}} du = \int \frac{u}{\sqrt{u ^{2} – 1}}\; du – \frac{1}{\sqrt{u ^{2} – 1}}\; du \\ For, \int \frac{u}{\sqrt{u ^{2} – 1}}\; du \\\) \(Suppose\; u ^{2} – 1 = z \\ 2u\; du = dz \\ Therefore,\; \int \frac{u}{\sqrt{u ^{2} – 1}}\; du = \frac{1}{2} \int \frac{dz}{\sqrt{z}} \\ = \frac{1}{2} \int z ^{- \frac{1}{2}} dz \\ = \frac{1}{2} [2 z^{\frac{1}{2}}] + C \\ = \sqrt{z} + C \\ = \sqrt{u ^{2} – 1} + C\) \(From\; the\; above\; equation\; we\; get \\ \int \frac{u – 1}{\sqrt{u ^{2} – 1}} du = \int \frac{u}{\sqrt{u ^{2} – 1}}\; du – \frac{1}{\sqrt{u ^{2} – 1}}\; du \\ = \sqrt{u ^{2} – 1} + C – log \left | u + \sqrt{u ^{2} – 1} \right | + C_{1} \\ = \sqrt{u ^{2} – 1} – log \left | u + \sqrt{u ^{2} – 1} \right | + (C + C_{1}) \\ = \sqrt{u ^{2} – 1} – log \left | u + \sqrt{u ^{2} – 1} \right | + C_{2}\)

Question 8:

Obtain an integral (or anti – derivative) of the \(\frac{u ^{2}}{\sqrt{u ^{6} + m ^{6}}}\)

Answer 8:

Suppose, u 3 = z

3 u2 du = dz

\(\int \frac{u ^{2}}{\sqrt{u ^{6} + m ^{6}}}\; du = \frac{1}{3} \int \frac{dz}{\sqrt{z ^{2} + (m^{3}) ^{2}}} \\ = \frac{1}{3} log \left | z + \sqrt{z ^{2} + m^{6}} \right | + C \\ = \frac{1}{3} log \left | u ^{3} + \sqrt{u ^{6} + m^{6}} \right | + C\)

Question 9:

Obtain an integral (or anti – derivative) of the \(\frac{sec ^{2}\; u}{\sqrt{tan ^{2} \;u + 4}}\)

Answer 9:

Suppose, tan u = z

sec 2 u du = dz

\(\int \frac{sec ^{2}\; u}{\sqrt{tan ^{2} \;u + 4}}\; du = \int \frac{dz}{\sqrt{z ^{2} + 2 ^{2}}} \\ = log \left | z + \sqrt{z ^{2} + 4} \right | + C \\ = log \left | tan\; u + \sqrt{tan ^{2} \;u + 4} \right | + C\)

Question 10:

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

Answer 10:

\(\int \frac{1}{\sqrt{u ^{2} + u + 2}} du = \int \frac{1}{(u + 1) ^{2} + (1) ^{2}} \\ Suppose,\; u + 1 = z \\ du = dz => \int \frac{1}{\sqrt{u ^{2} + 2 u + 2}} du = \int \frac{1}{\sqrt{z ^{2} + 1}} \;dz \\ = log \left | z + \sqrt{z ^{2} + 1} \right | + C \\ = log \left | (u + 1) + \sqrt{(u + 1) ^{2} + 1} \right | + C \\ = log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 1} \right | + C\)

Question 11:

Obtain an integral (or anti – derivative) of the \(\frac{1}{\sqrt{9 u ^{2} + 6 u + 2}}\)

Answer 11:

\(\int \frac{1}{\sqrt{9 u ^{2} + 6 u + 2}} du = \int \frac{1}{(3 u + 1) ^{2} + (2) ^{2}} \\ Suppose,\; 3 u + 1 = z \\ 3\; du = dz => \int \frac{1}{\sqrt{9 u ^{2} + 6 u + 2}} du = \frac{1}{3} \int \frac{1}{\sqrt{t ^{2} + 2 ^{2}}} \;dz \\ = \frac{1}{3} \left [ \frac{1}{2} tan ^{- 1} (\frac{z}{2}) \right ] + C \\ = \frac{1}{3} \left [ \frac{1}{2} tan ^{- 1} (\frac{3 u + 1}{2}) \right ] + C\)

Question 12:

Obtain an integral (or anti – derivative) of the \(\frac{1}{\sqrt{7 – 6 u – u ^{2}}}\)

Answer 12:

7 – 6 u – u2 = can also be written as 7 – (u2 + 6 u + 9 – 9)

Therefore,

7 – (u2 + 6 u + 9 – 9)

= 16 – (u2 + 6 u + 9)

= 16 – (u + 3) 2

= 4 2 – (u + 3) 2

\(\int \frac{1}{\sqrt{7 – 6 u – u ^{2}}} du = \int \frac{1}{4 ^{2} – (u + 3) ^{2}} du \\ Suppose,\; u + 3 = z \\ du = dz \\ \int \frac{1}{4 ^{2} – (u + 3) ^{2}} du = \int \frac{1}{4 ^{2} – (z) ^{2}} dz \\ = sin ^{- 1} (\frac{z}{4}) + C \\ = sin ^{- 1} (\frac{u + 3}{4}) + C\)

Question 13:

Obtain an integral (or anti – derivative) of the \(\frac{1}{\sqrt{(u – 1) (u – 2)}}\)

Answer 13:

(u – 1) (u – 2) can be written as u 2 – 3 u + 2

Therefore,

u 2 – 3 u + 2

\(= u ^{2} – 3 u + \frac{9}{4} – \frac{9}{4} + 2 \\ = \left ( u – \frac{3}{2} \right ) ^{2} – \frac{1}{4} \\ = \left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2} \\\) \(\int \frac{1}{\sqrt{(u – 1) (u – 2)}}\; du = \int \frac{1}{\sqrt{\left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2}}} \;du \\ Suppose,\; u – \frac{3}{2} = z \\ du = dz \\\) \(\int \frac{1}{\sqrt{\left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2}}} \;du = \int \frac{1}{\sqrt{z ^{2} – (\frac{1}{2}) ^{2}}} dz \\ = log \left | z + \sqrt{z ^{2} – (\frac{1}{2}) ^{2}} \right | + C \\ = log \left | (u – \frac{3}{2}) + \sqrt{(u – \frac{3}{2}) ^{2} – (\frac{1}{2}) ^{2}} \right | + C \\ = log \left | (u – \frac{3}{2}) + \sqrt{u^{2} – 3 u + 2} \right | + C\)

Question 14:

Obtain an integral (or anti – derivative) of the \(\frac{1}{\sqrt{8 + 3 u – u^{2}}}\)

Answer 14:

\(\frac{1}{\sqrt{8 + 3 u – u^{2}}} can\; also\; be\; written\; as\; 8 – \left ( u^{2} – 3 u + \frac{9}{4} – \frac{9}{4} \right ) \\ Therefore,\; \frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2} \\ \int \frac{1}{\sqrt{8 + 3 u – u^{2}}}\; du = \int \frac{1}{\sqrt{\frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2}}}\; du \\\) \(Suppose\; u – \frac{3}{2} = z \\ du = dz \\ \int \frac{1}{\sqrt{\frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2}}}\; du = \frac{1}{\sqrt{\left ( \frac{\sqrt{41}}{2} \right ) ^{2} – z ^{2} }} \; dz \\ = sin ^{- 1} \left ( \frac{z}{\frac{\sqrt{41}}{2}} \right ) + C \\ = sin ^{- 1} \left ( \frac{u – \frac{3}{2}}{\frac{\sqrt{41}}{2}} \right ) + C \\ = sin ^{- 1} \left ( \frac{2 u – 3}{\sqrt{41}} \right ) + C \\\)

Question 15:

Obtain an integral (or anti – derivative) of the \(\frac{1}{\sqrt{(u – m) (u – n)}}\)

Answer 15:

\((u – m) (u – n)\; can\; also\; be\; written\; as\; u ^{2} – (m + n) u + mn \\ Therefore,\; \\ u ^{2} – (m + n) u + mn \\ = u ^{2} – (m + n) u + \frac{(m + n) ^{2}}{4} – \frac{(m + n) ^{2}}{4} + mn \\ = \left [ u – \frac{(m + n)}{2} \right ] ^{2} – \frac{(m – n) ^{2}}{4} \\ \int \frac{1}{\sqrt{(u – m) (u – n)}}\; du = \int \frac{1}{\sqrt{\left \{u – \frac{(m + n)}{2} \right \} ^{2} – \left ( \frac{(m – n)}{2} \right ) ^{2}}} \;du \\\) \(Suppose,\; u – \left ( \frac{(m + n)}{2} \right ) = z \\ du = dz \\ \int \frac{1}{\sqrt{\left \{u – \frac{(m + n)}{2} \right \} ^{2} – \left ( \frac{(m – n)}{2} \right ) ^{2}}} \;du = \int \frac{1}{\sqrt{z ^{2} – \left ( \frac{m – n}{2} \right ) ^{2}}} \;dz \\\) \(log\; \left |z + \sqrt{z ^{2} – (\frac{m – n}{2}) ^{2}} \right | + C \\ log\; \left | \left \{ u – (\frac{m + n}{2}) \right \} + \sqrt{(u – m) (u – n)} \right | + C\)

Question 16:

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

Answer 16:

\(Suppose,\; 4u + 1 = A \frac{d}{dx} (2 u^{2} + u – 3) + B …. (1) \\ 4 u + 1 = A (4u + 1) + B \\ 4 u + 1 = 4 Au + A + B\)

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

4 A = 4 => A = 1

A + B = 1 => B = 0

From (1), we get

Suppose, 2 u2 + u – 3 = z

(4 u + 1) du = dz

\(\int \frac{4 u + 1}{\sqrt{2 u^{2} + u – 3}} du = \int \frac{1}{\sqrt{z}} \;dz \\ = 2 \sqrt{z} + C \\ = 2 \sqrt{2 u^{2} + u – 3} + C\)

Question 17:

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

Answer 17:

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

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

2 A = 1 => A = (1 / 2)

B = 2

From (1), we get

\(From\; (1), we\; get, \\ (u + 2) = \frac{1}{2} (2 u) + 2 \\ Then,\; \int \frac{u + 2}{\sqrt{u ^{2} – 1}}\; du = \int \frac{\frac{1}{2} (2 u) + 2}{\sqrt{u ^{2} – 1}} \\ = \frac{1}{2} \int \frac{2 u}{\sqrt{u ^{2} – 1}} \;du + \int \frac{2}{\sqrt{u ^{2} – 1}} \;du ….. (2) \\\) \(In\; \frac{1}{2} \int \frac{2 u}{\sqrt{u ^{2} – 1}} \;du = \frac{1}{2} \int \frac{dz}{\sqrt{z}} \\ = \frac{1}{2} [2 \sqrt{z}] + C \\ = \sqrt{z} + C \\ = \sqrt{u ^{2} – 1} \\ Then,\; \int \frac{2}{\sqrt{u ^{2} – 1}} \;du = 2 \int \frac{1}{\sqrt{u ^{2} – 1}} \;du \\ In\; equation\; (2), we\; get, \\ \int \frac{u + 2}{\sqrt{u ^{2} – 1}} \;du = \sqrt{u ^{2} – 1} + 2\; log\; \left | u + \sqrt{u ^{2} – 1} \right | + C\)

Question 18:

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

Answer 18:

Suppose, \(5 u – 2 = A \frac{d}{du} (1 + 2 u + 3 u^{2}) + B \\ 5 u – 2 = A (2 + 6 u) + B \\\)

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

\(5 = 6 A => A = \frac{5}{6} \\ 2 A + B = – 2 => B = – \frac{11}{3} \\ 5 u – 2 = \frac{5}{6} (2 + 6 u) + \frac{- 11}{3} \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \int \frac{\frac{5}{6} (2 + 6 u) – \frac{11}{3}}{1 + 2 u + 3 u^{2}} \;du \\ = \frac{5}{6} \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du – \frac{11}{3} \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\\) \(Suppose,\; I_{1} = \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du \; and\; I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} I_{1} – \frac{11}{3} I_{2} …. (1) \\ I_{1} = \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du \;and\; I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ Suppose,\; 1 + 2 u + 3 u^{2} = z \\ (2 + 6u)\; du = dz \\ I_{1} = \int \frac{dz}{z} \\ I_{1} = log \left | z \right | + C \\ I_{1} = log \left | 1 + 2 u + 3 u^{2} \right | + C ….. (2) I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\\) \(1 + 2 u + 3 u^{2} can\; also\; be\; written\; as\; 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u \right ) \\ Therefore, \; \\ 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u \right ) \\ = 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u + \frac{1}{9} – \frac{1}{9} \right ) \\ = 1 + 3 \left ( u + \frac{1}{3} \right ) ^{2} – \frac{1}{3} \\ = \frac{2}{3} + 3 \left ( u + \frac{1}{3} \right ) ^{2} \\ = 3 [\left ( u + \frac{1}{3} \right ) ^{2} + \frac{2}{9}] \\ = 3 [\left ( u + \frac{1}{3} \right ) ^{2} + (\frac{\sqrt{2}}{3}) ^{2}] \\\) \(I_{2} = \frac{1}{3} \int \left [ \frac{1}{[\left ( u + \frac{1}{3} \right ) ^{2} + (\frac{\sqrt{2}}{3}) ^{2}]} \right ] du \\ = \frac{1}{3} \left [ \frac{1}{\frac{\sqrt{2}}{3}}\; tan ^{- 1} (\frac{u + \frac{1}{3}}{\frac{\sqrt{2}}{3}}) \right ] \\ = \frac{1}{3} \left [ \frac{3}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) \right ] + C \\ = \frac{1}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) + C ….. (3) \\\)

Substituting equations (2) and (3) in equation (1), we get,

\(\int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du – \frac{11}{3} \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} \left [ log \left | 1 + 2 u + 3 u^{2} \right | \right ] – \frac{11}{3} \left [\frac{1}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) \right ] + C \\ = \frac{5}{6} log \left | 1 + 2 u + 3 u^{2} \right ] – \frac{11}{3 \sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) + C\)

Question 19:

Obtain an integral (or anti – derivative) of the \(\frac{6 u + 7}{\sqrt{(u – 5) (u – 4)}}\)

Answer 19:

Suppose, \(\frac{6 u + 7}{\sqrt{(u – 5) (u – 4)}} = \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \\ Suppose,\; 6 u + 7 = A \frac{d}{du} (u ^{2} – 9 u + 20) + B \\ 6 u + 7 = A (2 u – 9) + B\)

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

2 A = 6 => A = 3

– 9 + B = 7 => B = 34

6 u + 7 = 3 (2 u – 9) + 34

\(\int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = \int \frac{3 (2 u – 9) + 34}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ = 3 \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du + 34 \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du\) \(Suppose,\; I_{1} = \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du \;and \;I_{2} = \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ \int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = 3 I_{1} + 34 I_{2} \\ I_{1} = \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ Suppose,\; u ^{2} – 9 u + 20 = z (2 u – 9) \;du = dz \\ I_{1} = \frac{dz}{\sqrt{z}} \\ I_{1} = 2 \sqrt{z} \\ I_{1} = 2 \sqrt{u ^{2} – 9 u + 20} \\ and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ u ^{2} – 9 u + 20 \;can\; also \;be\; written\; as\; u ^{2} – 9 u + 20 + \frac{81}{4} – \frac{81}{4}\) \(Therefore,\; u ^{2} – 9 u + 20 + \frac{81}{4} – \frac{81}{4} \\ = \left ( u – \frac{9}{2} \right ) ^{2} – \frac{1}{4} \\ = \left ( u – \frac{9}{2} \right ) ^{2} – (\frac{1}{2}) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{\left ( u – \frac{9}{2} \right ) ^{2} – (\frac{1}{2}) ^{2}}} \;du \\ I_{2} = log \left |(u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right | + C ….. (3)\)

Substituting equations (2) and (3) in equation (1), we get,

\(\int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = 3 [2 \sqrt{u ^{2} – 9 u + 20}] + 34\; log \left [ \left ( (u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right ) \right ] + C \\ = 6 \sqrt{u ^{2} – 9 u + 20} + 34\; log \left [ \left ((u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right ) \right ] + C\)

Question 20:

Obtain an integral (or anti – derivative) of the \(\frac{u + 2}{\sqrt{4u – u ^{2}}}\)

Answer 20:

Suppose, \(u + 2 = A \frac{d}{du} (4u – u ^{2}) + B \\ (u + 2) = A (4 – 2 u) + B\)

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

\(– 2 A = 1 => A = – \frac{1}{2} \\ 4 A + B = 2 => B = 4 \\ (u + 2) = – \frac{1}{2} (4 – 2u) + 4 \\ \int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = \int \frac{- \frac{1}{2}\; (4 – 2u) + 4}{\sqrt{4 u – u^{2}}} \;du \\ = – \frac{1}{2} \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du + 4 \int \frac{1}{\sqrt{4 u – u^{2}}} \;du\) \(Suppose,\; I_{1} = \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du\; and\; I_{2} = \int \frac{1}{\sqrt{4 u – u^{2}}} \;du \\ \int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = – \frac{1}{2} I_{1} + 4 I_{2} ….. (1) \\ Then,\; I_{1} = \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du \\ Suppose,\; 4 u – u ^{2} = z \\ (4 – 2 u) \;du = dz \\ I_{1} = \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2\; \sqrt{4u – u ^{2}} ….. (2) \\ I_{2} = \int \frac{1}{\sqrt{4 u – u^{2}}} \;du \\ Suppose,\; 4 u – u^{2} = – (- 4 u + u^{2}) \\ (4 – 2 u) \;du = – (- 4 u + u^{2} + 4 – 4) \\ = 4 – (u – 2) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{(2) ^{2} – (u – 2) ^{2}}} \;du = sin ^{- 1} (\frac{u – 2}{2}) ….. (3) \\\)

Substituting equations (2) and (3) in equation (1), we get,

\(\int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = – \frac{1}{2} (2 \sqrt{4 u – u ^{2}}) + 4 sin^{- 1} \left ( \frac{u – 2}{2} \right ) + C \\ = – \sqrt{4u – u ^{2}} + 4 sin^{- 1} \left ( \frac{u – 2}{2} \right ) + C\)

Question 21:

Obtain an integral (or anti – derivative) of the \(\frac{u + 2}{\sqrt{4u ^{2} + 2 u + 3}}\)

Answer 21:

\(\int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} \int \frac{2 (u + 2)}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 4}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du + \frac{1}{2} \int \frac{2}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du + \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ Suppose,\; I_{1} = \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du \;and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ \int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} I_{1} + I_{2} \\\) \(I_{1} = \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ Suppose,\; u ^{2} + 2 u + 3 = z \\ (2 u + 2) \;du = dz \\ I_{1} = \int \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2 \sqrt{u ^{2} + 2 u + 3} ….. (2) \\ I_{2} = \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ u ^{2} + 2 u + 3 = u ^{2} + 2 u + 1 + 2 = (u + 1) ^{2} + (\sqrt{2}) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{(u + 1) ^{2} + (\sqrt{2}) ^{2}}} \;du = log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | ….. (3)\)

Substituting equations (2) and (3) in equation (1), we get,

\(\int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} (2 \sqrt{u ^{2} + 2 u + 3}) + log\; \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | + C \\ = \sqrt{u ^{2} + 2 u + 3} + log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | + C\)

Question 22:

Obtain an integral (or anti – derivative) of the \(\frac{u + 2}{\sqrt{u ^{2} – 2 u – 5}}\)

Answer 22:

\(Suppose,\; (u + 3) = A \frac{d}{du} (u ^{2} – 2u – 5) \\ (u + 3) = A (2 u – 2) + B \\\)

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

\(2 A = 1 => A = \frac{1}{2} \\ – 2 A + B = 3 => B = 4 \\ (u + 3) = \frac{1}{2} (2 u – 2) + 4 \\ \int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \int \frac{\frac{1}{2} (2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \\ = \frac{1}{2} \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du + 4 \int \frac{1}{u ^{2} – 2 u – 5} du \\\) \(Suppose,\; I_{1} = \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \;and\; I_{2} = \int \frac{1}{u ^{2} – 2 u – 5} du \\ \int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \frac{1}{2} I_{1} + 4 I_{2} …. (1) \\ Then, I_{1} = \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \\ Suppose,\; u ^{2} – 2 u – 5 = Z \\ (2 u – 2) \;du = dz \\ I_{1} = \int \frac{dz}{z} \\ = log \left | z \right | + C \\ = log \left | u ^{2} – 2 u – 5 \right | + C …. (2)\) \(I_{2} = \int \frac{1}{u ^{2} – 2 u – 5} du \\ = \int \frac{1}{(u ^{2} – 2 u + 1) – 6} du \\ = \int \frac{1}{(u – 1) ^{2} + (\sqrt{6}) ^{2}} \;du \\ = \frac{1}{2 \sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right ) …… (3)\)

Using equations (2) and (3) in equation (1), we get,

\(\int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \frac{1}{2} log \left | u ^{2} – 2 u – 5 \right | + 4 \left [ \frac{1}{2 \sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right ) \right ] + C \\ = \frac{1}{2} log \left | u ^{2} – 2 u – 5 \right | + \frac{2}{\sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right )\)

Question 23:

Obtain an integral (or anti – derivative) of the \(\frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}}\)

Answer 23:

\(Suppose,\; 5u + 3 = A \frac{d}{du} (u ^{2} + 4 u + 10) + B \\ 5u + 3 = A (2 u + 4) + B\)

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

\(2 A = 5 => A = \frac{5}{2} \\ 4 A + B = 3 => B = – 7 \\ 5 u + 3 = \frac{5}{2} (2 u + 4) – 7 \\ \int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{\frac{5}{2} ((2 u + 4) – 7)}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ = \frac{5}{2} \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du – 7 \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ Suppose,\; I_{1} = \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du \;and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ \int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{5}{2} I_{1} – 7 I_{2} \;\; ….. (1)\) \(I_{1} = \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ Suppose,\; u ^{2} + 4 u + 10 = z \\ (2 u + 4) \;du = dz \\ I_{1} = \int \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2 \sqrt{u ^{2} + 4 u + 10} ….. (2) I_{2} = \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ = \int \frac{1}{\sqrt{(u ^{2} + 4 u + 4) + 6}} \;du \\ = \int \frac{1}{(u + 2) ^{2} + (\sqrt{6}) ^{2}} \;du \\ =log \left |(u + 2)\sqrt{u ^{2} + 4 u + 10} \right | ….. (3)\)

Substituting equations (2) and (3) in equation (1), we get,

\(\int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{5}{2} [2 \sqrt{u ^{2} + 4 u + 10}] – 7 log \left | (u + 2) + \sqrt{u ^{2} + 4 u + 10} \right | + C \\ = 5\; \sqrt{u ^{2} + 4 u + 10} – 7 log \left | (u + 2) + \sqrt{u ^{2} + 4 u + 10} \right | + C \\\)

Question 24: Which of the following below is the answer for \(\int \frac{du}{u ^{2} + 2 u + 2} \;du\)

\((a) u tan ^{- 1} (u + 1) + C \\ (b) tan ^{- 1} (u + 1) + C \\ (c) (u + 1) tan ^{- 1} (u) + C \\ (d) tan ^{- 1} (u) + C\)

Answer 24:

\(\int \frac{du}{u ^{2} + 2 u + 2} \;du = \int \frac{du}{(u ^{2} + 2 u + 1) + 1} \\ = \int \frac{1}{(u + 1) ^{2} + (1) ^{2}} \;du \\ = \left [ tan ^{- 1} (u + 1) \right ] + C\)

Thus, (b) is the correct answer.

Question 25: Which of the following below is the answer for \(\int \frac{du}{\sqrt{9 u – 4 u ^{2}}} \;du\)

\((a) \frac{1}{9} sin ^{- 1} \frac{9 u – 8}{8} + C \\ (b) \frac{1}{2} sin ^{- 1} \frac{8 u – 9}{9} + C \\ (c) \frac{1}{3} sin ^{- 1} \frac{9 u – 8}{8} + C \\ (d) \frac{1}{2} sin ^{- 1} \frac{9 u – 8}{8} + C\)

Answer 25:

\(\int \frac{du}{\sqrt{9 u – 4 u ^{2}}} \;du \\ = \int \frac{du}{\sqrt{- 4 \left ( u ^{2} – \frac{9}{4} u \right )}} \\ = \int \frac{du}{\sqrt{- 4 \left ( u ^{2} – \frac{9}{4} u + \frac{81}{64} – \frac{81}{64} \right )}} \\ = \int \frac{1}{\sqrt{- 4} \left [ \left ( u – \frac{9}{8} \right ) ^{2} – (\frac{9}{8}) ^{2} \right ]} \;du \\ = \frac{1}{2} \int \frac{1}{\sqrt{(\frac{9}{8}) ^{2} – \left ( u – \frac{9}{8} \right ) ^{2}}} \;du \\ = \frac{1}{2} \left [ sin ^{- 1} \left ( \frac{u – \frac{9}{8}}{\frac{9}{8}} \right ) \right ] + C \\ = \frac{1}{2} sin ^{- 1} \left ( \frac{8 u – 9}{9} \right ) + C\)

Thus, (b) is the correct answer.