Question

\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then 

Options

• \[ a = \frac{1}{3}, b = 1\]

• \[a = - \frac{1}{3}, b = 1\]

• \[ a = - \frac{1}{3}, b = - 1\]

• \[ a = \frac{1}{3}, b = - 1\]

   

Answer 1

\[ a = \frac{1}{3}, b = - 1\]

 

\[\text{Let }I = \int\frac{x^3}{\sqrt{1 + x^2}}dx\]

\[ = \int\frac{x . x^2}{\sqrt{1 + x^2}}dx\]

\[\text{Let }\left( 1 + x^2 \right) = t\]

\[\text{On differentiating both sides, we get}\]

\[ 2x\ dx = dt\]

\[ \therefore I = \frac{1}{2}\int\frac{t - 1}{\sqrt{t}}dt\]

\[ = \frac{1}{2}\int\left( \frac{t}{\sqrt{t}} - \frac{1}{\sqrt{t}} \right)dt\]

\[ = \frac{1}{2}\int\left( t^\frac{1}{2} - t^\frac{- 1}{2} \right)dt\]

\[ = \frac{1}{2}\left( \frac{2}{3} t^\frac{3}{2} - \frac{2}{1} t^\frac{1}{2} \right) + C\]

\[ = \left( \frac{1}{3} \left( 1 + x^2 \right)^\frac{3}{2} - \sqrt{1 + x^2} \right) + C\]

\[\text{Since, }\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\]

\[\text{Therefore, }a = \frac{1}{3}, b = - 1 . \]