Integrating $\int_{0}^{1} \frac{\ln(1-x)}{x^2 + 1}$ via substitution and then Feynman's Technique I've been trying the find a solution to the post: Integrating $\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx$
And I was wondering if the substitution $u = -x$, so that the bounds become $u(0) = 0$ and $u(1) = -1$, would be possible. 
This would make solving the integral possible, as I've already seen that the integral, 
$$ \int_{0}^{1} \frac{\ln(1 + x)}{x^2 + 1} dx $$
can be solved using Feynman's technique. Link: https://www.youtube.com/watch?v=b4ZuFeInVHE
This substitution would be very nice, but I'm sceptical to use it as the bounds seem to become nonsense. 
 A: Following Cornel's method in this solution ( check the integral $\mathcal{J})$
$$I=\int_0^1\frac{\ln(1-x)}{1+x^2}\ dx =\Im\int_0^1\frac{i\ln(1-x)}{1-ix}\ dx\overset{1-x=t}{=}\Im\int_0^1\frac{i\ln(t)}{1-i+it}\ dt$$
$$=\Im\text{Li}_2\left(\frac{i}{i-1}\right)=\Im\left(-\text{Li}_2(i)-\frac12\ln^2(1-i)\right)=\frac{\pi}8\ln(2)-G$$
where in the last step, I used landen's identity.
A: \begin{aligned} \int_{0}^{1}{\frac{\ln{\left(1-x\right)}}{1+x^{2}}\,\mathrm{d}x}=\int_{0}^{\frac{\pi}{4}}{\ln{\left(1-\tan{x}\right)}\,\mathrm{d}x}&=\int_{0}^{\frac{\pi}{4}}{\ln{\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}-\int_{0}^{\frac{\pi}{4}}{\ln{\left(\cos{x}\right)}\,\mathrm{d}x}\\ &=\int_{0}^{\frac{\pi}{4}}{\ln{\left(\sqrt{2}\sin{\left(\frac{\pi}{4}-x\right)}\right)}\,\mathrm{d}x}-\int_{0}^{\frac{\pi}{4}}{\ln{\left(\cos{x}\right)}\,\mathrm{d}x}\\ &=\frac{\pi}{8}\ln{2}+\int_{0}^{\frac{\pi}{4}}{\ln{\left(\sin{x}\right)}\,\mathrm{d}x}-\int_{0}^{\frac{\pi}{4}}{\ln{\left(\cos{x}\right)}\,\mathrm{d}x}\\ &=\frac{\pi}{8}\ln{2}+\int_{0}^{\frac{\pi}{4}}{\ln{\left(\tan{x}\right)}\,\mathrm{d}x} \\ \int_{0}^{1}{\frac{\ln{\left(1-x\right)}}{1+x^{2}}\,\mathrm{d}x}&=\frac{\pi}{8}\ln{2}-G\end{aligned}
Where $ G $ is Catalan's constant.
A: Different approach using harmonic series.
$$I=\int_0^1\frac{\ln(1-x)}{1+x^2}\ dx=\sum_{n=0}^\infty(-1)^n\int_0^1 x^{2n}\ln(1-x)\ dx=-\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{2n+1}$$
Use the fact that 
$$\sum_{n=0}^\infty (-1)^n f(2n+1)=\Im\sum_{n=1}^\infty i^nf(n)$$
$$\Longrightarrow I=-\Im\sum_{n=1}^\infty\frac{i^nH_n}{n}=-\Im\left(\text{Li}_2(i)+\frac12\ln^2(1-i)\right)=\frac{\pi}{8}\ln(2)-G$$
A: Something different that the integral you've to calculate.
If you wan't to use Feynman technic :
You can try $$I(a)=\int _0^1\frac{\ln \left(1-ax\right)}{x^2+1}\:dx$$ $$a<1$$ 
Suppose you can use Leibgniz theorem,
$$ I'(a)=\int _0^1\frac{-a\arctan(x)}{1-ax}\:dx $$
Now use integration by part on that integral : 
$$ I'(a)=\int _0^1\frac{-a\arctan(x)}{1-ax}\:dx = [-a\ln(1-ax)\arctan(x)]_0^1  + a\int_0^1I(a)$$
You've a differential equation :
$$ I'(a)=-a\ln(1-a)\frac{\pi}{4}+aI(a)$$
You can solve it know and use value of $I(0)$ for initial conditions, simple to calculate.
