Integrating $\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx$ When i came across this one i thought of using the substitution $x=\frac{1-t}{1+t}$ that lead to an easy expression.
$$\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx\overset{x=\frac{1-t}{1+t}}=\ln \left(2\right)\int _0^1\frac{1}{t^2+1}\:dt+\int _0^1\frac{\ln \left(t\right)}{t^2+1}\:dt-\int _0^1\frac{\ln \left(t+1\right)}{t^2+1}\:dt$$
$$=\frac{\pi }{4}\ln \left(2\right)-G-\frac{\pi }{8}\ln \left(2\right)=\boxed{\frac{\pi }{8}\ln \left(2\right)-G}$$
But then i tried to solve it using feynman's trick with the following parameter:
$$I=\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx$$
$$I\left(a\right)=\int _0^1\frac{\ln \left(1-ax\right)}{x^2+1}\:dx$$
$$I'\left(a\right)=-\int _0^1\frac{x}{\left(x^2+1\right)\left(1-ax\right)}\:dx=\frac{1}{a^2+1}\int _0^1\frac{a-x}{\left(x^2+1\right)}-\frac{a}{1-ax}\:dx$$
$$=\frac{1}{a^2+1}\left(\frac{\pi a}{4}-\frac{\ln \left(2\right)}{2}+\ln \left(1-a\right)\right)\:$$
$$\int _0^1I'\left(a\right)\:da=\frac{\pi }{4}\int _0^1\frac{a}{a^2+1}\:da-\frac{\ln \left(2\right)}{2}\int _0^1\frac{1}{a^2+1}\:da+\int _0^1\frac{\ln \left(1-a\right)}{a^2+1}\:da\:$$
$$I=\frac{\pi }{8}\ln \left(2\right)-\frac{\pi }{8}\ln \left(2\right)+I$$
And everything cancels.
¿Am i doing something wrong?, ¿Should i be using another parameter or other upper and lower bounds?

You are not allowed to switch the integral and derivative. The conditions of the theorem about switching those are not met. Denote $\frac{\ln(1-ax)}{x^2+1}$ by $f(x,a)$, one of the conditions is that  there must exist a function $g:[0,1] \rightarrow \mathbb{C} $ such that $\int_0^1 |g(x)|dx < \infty$ and $| \frac{\partial f}{\partial a}(x,a) |\leq g(x)$ for all $a$. 
