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 $$\sum_{n=0}^\infty\frac{(-1)^n H_{2n+1}}{2n+1}=-\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x.$$
 A: Another chance:
$$\begin{eqnarray*} \int_{0}^{1}\arctan(x)\left(\frac{1}{x}-\frac{1}{x+1}\right)\,dx &\stackrel{IBP}{=}&\left[\log\left(\frac{x}{x+1}\right)\arctan(x)\right]_{0}^{1}-\int_{0}^{1}\frac{\log\left(\frac{x}{x+1}\right)}{x^2+1}\,dx\\&=&-\frac{\pi}{4}\log 2+K+\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\,dx\end{eqnarray*} $$
so the claim is equivalent to
$$ \int_{0}^{1}\frac{\log(1-x^2)}{1+x^2}\,dx = \frac{\pi}{4}\log(2)-K \tag{1}$$
with $K$ being Catalan's constant. $(1)$ is well-suited for the application of Fourier series. Indeed, up to the substitution $x=\tan\theta$ the LHS of $(1)$ becomes
$$ \int_{0}^{\pi/4}\left(\log\cos(2\theta)-2\log\sin(\theta)\right)\,d\theta \tag{2}$$
which is simple to tackle through the Fourier series of $\log\sin$ and $\log\cos$.
A: 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.
