Answer
$$\frac{d}{d x}\left(\frac{1}{2} \tan ^{-1} x+\frac{1}{4} \ln \frac{(x+1)^{2}}{x^{2}+1}\right) =\frac{1}{\left(x+1\right)\left(x^2+1\right)}$$
Work Step by Step
Since
\begin{align*}
\frac{d}{d x}\left(\frac{1}{2} \tan ^{-1} x+\frac{1}{4} \ln \frac{(x+1)^{2}}{x^{2}+1}\right)&= \frac{d}{dx}\left(\frac{1}{2}\tan^{-1} \left(x\right)\right)+\frac{d}{dx}\left(\frac{1}{4}[2\ln (x+1)-\ln(x^2+1) \right)\\
&=\frac{1}{2(1+x^2)}+\frac{1}{2(x+1)}-\frac{x}{2(x^2+1)}\\
&=\frac{1}{\left(x+1\right)\left(x^2+1\right)}
\end{align*}
