Answer
$$x\tan^{-1}x-x+\tan^{-1}x+c$$
Work Step by Step
Given $$\int\tan^{-1}\sqrt{x}dx$$
Let $u^2=x \ \ \ \Rightarrow \ \ 2udu=dx$, then
$$\int\tan^{-1}\sqrt{x}dx= 2\int u\tan^{-1}udu$$
Let
\begin{align*}
u&= \tan^{-1}u\ \ \ \ \ \ \ \ \ \ \ \ dv=2udu\\
du&=\frac{1}{1+u^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v= u^2
\end{align*}
Then
\begin{align*}
\int\tan^{-1}\sqrt{x}dx&=\int\tan^{-1}\sqrt{x}dx\\
&=u^2 \tan^{-1}u-\int \frac{u^2}{1+u^2}du\\
&=u^2 \tan^{-1}u-\int \frac{u^2+1-1}{1+u^2}du\\
&=u^2 \tan^{-1}u -\int 1-\frac{ 1}{1+u^2}du\\
&=u^2 \tan^{-1}u-u+\tan^{-1}u+c\\
&=x\tan^{-1}x-x+\tan^{-1}x+c
\end{align*}