Question

$$\begin{gathered} \int {\tan }^{-1}\sqrt{x}dxisequalto \\ \left(a\right)(x+1){\tan }^{-1}\sqrt{x}-\sqrt{x}+C\left(b\right)x{\tan }^{-1}\sqrt{x}-\sqrt{x}+C \\ \left(c\right)\sqrt{x}-x{\tan }^{-1}\sqrt{x}+C\left(d\right)\sqrt{x}-(x+1){\tan }^{-1}\sqrt{x}+C \end{gathered}$$

Answer 1

$I=\int {\tan }^{-1}\sqrt{x}dx$

Put $x={\tan }^2\theta$

$\Rightarrow dx=2\tan \theta {\sec }^2\theta d\theta$

$\therefore I=\int {\tan }^{-1}\sqrt{{\tan }^2\theta }\times 2\tan \theta {\sec }^2\theta d\theta$

$\Rightarrow I=2\int \theta \times \tan \theta {\sec }^2\theta d\theta$

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

$\Rightarrow I=2\left[\theta \times \frac{{\tan }^2\theta }{2}-\int \frac{{\tan }^2\theta }{2}d\theta \right]$

$\Rightarrow I=\theta {\tan }^2\theta -\int \left({\sec }^2\theta -1\right)d\theta$

$\Rightarrow I=\theta {\tan }^2\theta -\int {\sec }^2\theta d\theta +\int d\theta$

$\Rightarrow I=\theta {\tan }^2\theta -\tan \theta +\theta +C$

$\Rightarrow I={\tan }^{-1}\sqrt{x}\times x-\sqrt{x}+{\tan }^{-1}\sqrt{x}+C$ $\left(x={\tan }^2\theta \Rightarrow \tan \theta =\sqrt{x}\Rightarrow \theta ={\tan }^{-1}\sqrt{x}\right)$

$\Rightarrow I=\left(x+1\right){\tan }^{-1}\sqrt{x}-\sqrt{x}+C$

Hence, the correct answer is option (a).