Question

If$\int_{}^{}{x^{–1/2}(2 + 3x^{1/3})^{- 2}dx}$ = A tan^–1$\left\{ \sqrt{\frac{3}{2}}x^{1/6} \right\}$+ B $\frac{x^{1/6}}{2 + 3x^{1/3}}$+ C then –

• A. A = $\frac{1}{6}$, B = 6
• B. A = B = $\frac{1}{\sqrt{6}}$
• C. A = $\frac{1}{\sqrt{6}}$, B = –$\frac{1}{\sqrt{6}}$
• D. A = $\frac{1}{\sqrt{6}}$, B = –1

Answer 1

Answer: (D)

A = $\frac{1}{\sqrt{6}}$, B = –1

Answer 2

Let I = $\int_{}^{}x^{- 1/2}$ (2 + 3x^1/3)^–2 dx

Put x = t⁶ \ dx = 6t⁵dt

then I = $\int_{}^{}t^{- 3}$ (2 + 3t²)^–2 . 6t⁵ dt = 6

$\int_{}^{}\frac{t^{2}}{(2 + 3t^{2})^{2}}$dt =$\frac { 6 } { 9 }$ $\int_{}^{}\frac{t^{2}dt}{\left( \frac{2}{3} + t^{2} \right)}$

Now put t = $\sqrt{\left( \frac{2}{3} \right)}$tan q

\ dt = $\sqrt{\left( \frac{2}{3} \right)}$sec² q dq

\ I = $\frac { 6 } { 9 }$ $\int_{}^{}\frac{\frac{2}{3}\tan^{2}\theta.\sqrt{\left( \frac{2}{3} \right)}\sec^{2}\theta d\theta}{\frac{4}{9}\sec^{4}\theta}$

= $\sqrt { \frac { 2 } { 3 } }$ $\int_{}^{}{\sin^{2}\theta d\theta}$

= $\frac{1}{\sqrt{6}}$ $\int_{}^{}{(1 - \cos 2\theta)d\theta}$= $\frac { 1 } { \sqrt { 6 } }$ $\left\{ \theta - \frac{\sin 2\theta}{2} \right\}$ + c

= $\frac { 1 } { \sqrt { 6 } }$ $\left\{ \theta - \frac{\tan\theta}{1 + \tan^{2}\theta} \right\}$+ c

= $\frac { 1 } { \sqrt { 6 } }$ $\left\{ \tan^{- 1}\left\{ \sqrt{\frac{3}{2}}t \right\} - \frac{\sqrt{\frac{3}{2}}.t}{1 + \frac{3}{2}t^{2}} \right\}$ + c

= $\frac { 1 } { \sqrt { 6 } }$ $\left\{ \tan^{- 1}\left\{ \sqrt{\frac{3}{2}}x^{1/6} \right\} - \frac{\sqrt{6}x^{1/6}}{2 + 3x^{1/3}} \right\}$

+ c$\left\{ \because\tan\theta = \sqrt{\frac{3}{2}}t \right\}$