Answer
$$\frac{\pi^{2}}{4}$$
Work Step by Step
Given
$$\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x $$
Since by Exercise 64
$$ \int_{0}^{\pi} x f(\sin x) d x =\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) d x $$
we get
\begin{aligned}
\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x&=\int_{0}^{\pi} x f(\sin x) d x\\
&=\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) d x\\
&=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x
\end{aligned}
Let $ u=\cos x$, then $du=-\sin xdx$ and
\begin{aligned}
\text{at } \ x&= 0\ \ \to u = 1 \\
\text{at } \ x&=\pi\ \ \to u = -1
\end{aligned}
It follows that
\begin{aligned}
\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x &=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x\\
&= -\frac{\pi}{2} \int_{1}^{-1} \frac{d u}{1+u^{2}}\\
&=\frac{\pi}{2} \int_{-1}^{1} \frac{d u}{1+u^{2}}\\
&=\frac{\pi}{2}\left[\tan ^{-1} u\right]_{-1}^{1}\\
&=\frac{\pi}{2}\left[\tan ^{-1} 1-\tan ^{-1}(-1)\right]\\
&=\frac{\pi}{2}\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]\\
&=\frac{\pi^{2}}{4}
\end{aligned}
