Question

\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int \frac{x \tan^{- 1} x}{\left( 1 + x^2 \right)^\frac{3}{2}}\text{ dx }\]

\[\text{ Putting x }= \tan \theta\]

\[ \Rightarrow dx = \sec^2  \text{ θ dθ }\]

\[\text{and}\ \theta = \tan^{- 1} x\]

\[ \therefore I = \int \frac{\left( \tan \theta \right) . \theta . \sec^2   \text{ θ dθ }}{\left( 1 + \tan^2 \theta \right)^\frac{3}{2}}\]

\[ = \int \frac{\theta . \tan \theta \sec^2   \text{ θ dθ }}{\left( \sec^2 \theta \right)^\frac{3}{2}}\]

\[ = \int \frac{\theta \tan \theta . \sec^2   \text{ θ dθ }}{\sec^3 \theta}\]

\[ = \int \frac{\theta . \tan \theta}{\sec \theta} d\theta\]

\[ = \int \theta_I . \sin_{II} \text{ θ dθ }\]

\[ = \theta\int\sin \text{ θ dθ }\] - \int\left\{ \frac{d}{d\theta}\left( \theta \right)\int\sin d\theta \right\}d\theta\]

\[ = \theta \left( - \cos \theta \right) - \int1 . \left( - \cos \theta \right) d\theta\]

\[ = - \theta \cos \theta + \sin \theta + C\]

\[ = \frac{- \theta}{\sec \theta} + \frac{1}{cosec   \text{ θ }} + C\]

\[ = \frac{- \theta}{\sqrt{1 + \tan^2 \theta}} + \frac{1}{\sqrt{1 + \cot^2 \theta}} + C\]

\[ = \frac{- \theta}{\sqrt{1 + \tan^2 \theta}} + \frac{\tan \theta}{\sqrt{\tan^2 \theta + 1}} + C\]

\[ = \frac{- \tan^{- 1} x}{\sqrt{1 + x^2}} + \frac{x}{\sqrt{x^2 + 1}} + C\]