<s>[INST] Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$ How to show that
$$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$$
without evaluating each integral individually?
I came up with this problem while working on some problem where I found ( by comparing some results) that
$$\sum_{n=0}^\infty\frac{(-1)^n H_{2n+1}}{2n+1}=\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x$$
and by using $$\int_0^1 x^{2n}\ln(1-x)
\mathrm{d}x=-\frac{H_{2n+1}}{2n+1},$$
we have $$\sum_{n=0}^\infty\frac{(-1)^n H_{2n+1}}{2n+1}=-\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x.$$
 [/INST] Performing integration by parts,
\begin{align*}
\int_{0}^{1} \frac{\log(1-x)}{1+x^2} \, \mathrm{d}x
&= \underbrace{\left[ \log(1-x) \left(\arctan x - \frac{\pi}{4}\right) \right]_{0}^{1}}_{=0} - \int_{0}^{1} \frac{\frac{\pi}{4} - \arctan x}{1 - x} \, \mathrm{d}x.
\end{align*}
Now substituting $x = \frac{1-t}{1+t}$ and using the identity $\arctan\bigl(\frac{1-t}{1+t}\bigr) = \frac{\pi}{4} - \arctan t$ for $t > -1$, we get
\begin{align*}
\int_{0}^{1} \frac{\frac{\pi}{4} - \arctan x}{1 - x} \, \mathrm{d}x
&= \int_{0}^{1} \frac{\arctan t}{t(t+1)} \, \mathrm{d}t.
\end{align*}
This proves the desired identity.
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