Answer
$$\frac{\pi }{4}-\frac{1}{2}\ln \left(2\right)$$
Work Step by Step
Given $$\int_{0}^{1} \tan ^{-1} x d x$$
Let
\begin{align*}
u&= \tan^{-1}x\ \ \ \ \ \ \ dv= dx\\
du&=\frac{1}{1+x^2}dx\ \ \ \ \ \ v=x
\end{align*}
Then
\begin{align*}
\int_{0}^{1} \tan ^{-1} x d x&= x\tan^{-1}x\bigg|_{0}^{1}-\int_{0}^{1} \frac{x}{1+x^2}dx\\
&= x\tan^{-1}x-\frac{1}{2}\ln (1+x^2)\bigg|_{0}^{1}\\
&=\frac{\pi }{4}-\frac{1}{2}\ln \left(2\right)
\end{align*}
