Question

$$\int {\tan }^{-1}\left(\sqrt{x}\right)\text{dx }$$

Answer 1

$$\text{ Let I }=\int {\tan }^{-1}\sqrt{x}\text{ dx }$$
$$=\int \frac{\sqrt{x}.{\tan }^{-1}\sqrt{x}\text{ dx}}{\sqrt{x}}$$
$$\text{ Let }\sqrt{x}=t$$
$$\Rightarrow \frac{1}{2\sqrt{x}}\text{ dx }=dt$$
$$=\frac{dx}{\sqrt{x}}=2dt$$
$$\therefore I=2\int t_{II}.{\tan }_I^{-1}\left(t\right)\text{ dt }$$
$$=2[{\tan }^{-1}t\int \text{ t dt }-\int \{\frac{d}{dt}({\tan }^{-1}t)\int \text{ t dt }\}dt]$$
$$=2[{\tan }^{-1}\left(t\right).\frac{t^2}{2}-\int \frac{1}{1+t^2}.\frac{t^2}{2}dt]$$
$$={\tan }^{-1}\left(t\right).t^2-\int \frac{t^2}{1+t^2}dt$$
$$={\tan }^{-1}\left(t\right).t^2-\int \left(\frac{1+t^2-1}{1+t^2}\right)dt$$
$$={\tan }^{-1}\left(t\right).t^2-\int dt+\int \frac{dt}{1+t^2}$$
$$={\tan }^{-1}\left(t\right).t^2-t+{\tan }^{-1}\left(t\right)+C(\because \sqrt{x}=t)$$
$$={\tan }^{-1}\left(\sqrt{x}\right).x-\sqrt{x}+{\tan }^{-1}\sqrt{x}+C$$
$$=(x+1){\tan }^{-1}\sqrt{x}-\sqrt{x}+C$$