Question

\[\int \tan^{- 1} \sqrt{x}\ dx\]

Answer 1

\[\text{We have}, \]

\[I = \int \tan^{- 1} \sqrt{x}\text{  dx }\]

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

\[ \Rightarrow x = \tan^2 \theta\]

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

\[ \therefore I = \int \left[ \tan^{- 1} \left( \tan \theta \right)\text{  2   tan  θ } \sec^\text{ 2 }\theta \right] d\theta\]

\[ = 2 \int \theta_i  \text{ tan  θ  sec}^2_{ii}   \text{ θ   dθ }\]

\[ = 2\left[ \theta \times \frac{\tan^2 \theta}{2} - \int1\frac{\tan^2 \text{ θ   dθ }}{2} \right] ................\left( \because \int \tan \theta \sec^2 \text{ θ   dθ } = \frac{\tan^2 \theta}{2} \right)\]

\[ = 2\left[ \theta\frac{\tan^2 \theta}{2} - \frac{1}{2}\int\left( \sec^2 \theta - 1 \right)d\theta \right]\]

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

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

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