Integral $ \int_0^\infty \frac{\ln x}{(x+c)(x-1)} dx$ I've been trying to solve the following integral for days now.
$$P = \int_0^\infty \frac{\ln(x)}{(x+c)(x-1)} dx$$
with $c > 0$. I figured out (numerically, by accident) that if $c = 1$, then $P = \pi^2/4$. But why? And more importantly: what's the general solution of $P$, for given $c$? I tried partial fraction expansions, Taylor polynomials for $ln(x)$ and more, but nothing seems to work. I can't even figure out where the $\pi^2/4$ comes from.
(Background: for a hobby project I'm building a machine learning algorithm that predicts sports match scores. Somehow the breaking point is this integral, so solving it would get things moving again.)
 A: A more general way:
Use the classic integral
$$\int_0^\infty \frac{x^p}{a+x}\;dx=-a^p\frac{\pi}{\sin(\pi p)};-1<p<0$$
Then
$$\int_0^\infty\frac{x^p}{(a+x)(c+x)}\;dx=\frac{\pi}{\sin(\pi p)}\frac{c^p-a^p}{c-a}$$
Now differentiate this  with respect to $p$ and compute the limit of the result of differentiation as $p$ approaches $0$ to get
$$\int_0^\infty\frac{\ln x}{(a+x)(c+x)}\;dx=\frac{1}{2}\frac{\ln^2 c-\ln^2 a}{c-a}$$
A: $$\bbox[10pt, border:2px, lightblue]{\int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx=\frac{\pi^2+\ln^2 c}{2(1+c)},\ \ c>0}$$
A  nice solution can be found here due to Yaghoub Sharifi.

Perhaps it might be into your interest to see a solution for the following integral:
$$I(a,b)=\int_0^\infty \frac{\ln x}{(x+a)(x+b)}dx\overset{x\to \frac{ab}{x}}=\int_0^\infty \frac{\ln\left(\frac{ab}{x}\right)}{(x+a)(x+b)}dx$$
Summing up the two integrals from above gives:
$$2I(a,b)=\ln(ab)\int_0^\infty \frac{1}{(x+a)(x+b)}dx\Rightarrow \boxed{I(a,b)=\frac{\ln(ab)}{2}\frac{\ln\left(\frac{a}{b}\right)}{a-b},\ \ a,b>0}$$
One might force putting $a=c, b=-1$ in the above and take $\ln(-1)=i\pi$ (the principal value).
$$\Rightarrow I(c,-1)=\frac{\ln^2 c-\ln^2 (-1)}{2(c+1)}=\frac{\pi^2 +\ln^2 c}{2(1+c)}$$
A: A different approach using polylogarithms
For this solution the following identities are used

$\displaystyle\text{Li}_2\left(z\right)=-\int _0^z\frac{\ln \left(1-t\right)}{t}\:dt$, $\displaystyle \int _0^1\frac{c\ln ^n\left(x\right)}{1-cx}\:dx=\left(-1\right)^nn!\text{Li}_{n+1}\left(c\right)$, $\displaystyle \text{Li}_2\left(-z\right)+\text{Li}_2\left(-\frac{1}{z}\right)=-\zeta \left(2\right)-\frac{1}{2}\ln ^2\left(z\right)$

$$\int_0^{\infty}\frac{\ln\left(x\right)}{\left(c+x\right)\left(x-1\right)}\:dx$$
$$=\int _0^1\frac{\ln \left(x\right)}{\left(c+x\right)\left(x-1\right)}\:dx+\underbrace{\int _1^{\infty }\frac{\ln \left(x\right)}{\left(c+x\right)\left(x-1\right)}\:dx}_{x=\frac{1}{x}}$$
$$\hspace{-5mm}=\frac{2}{1+c}\int _0^1\frac{\ln \left(x\right)}{x-1}\:dx-\frac{1}{1+c}\underbrace{\int _0^1\frac{\ln \left(x\right)}{c+x}\:dx}_{x=ct}-\frac{c}{1+c}\int _0^1\frac{\ln \left(x\right)}{1+cx}\:dx$$
$$\hspace{-8mm}=\frac{2}{1+c}\zeta \left(2\right)-\frac{\ln \left(c\right)}{1+c}\int _0^{\frac{1}{c}}\frac{1}{1+t}\:dt-\frac{1}{1+c}\underbrace{\int _0^{\frac{1}{c}}\frac{\ln \left(t\right)}{1+t}\:dt}_{\text{IBP}}-\frac{1}{1+c}\text{Li}_2\left(-c\right)$$
$$\hspace{-2mm}=\frac{2}{1+c}\zeta \left(2\right)-\frac{\ln \left(c\right)}{1+c}\ln \left(1+\frac{1}{c}\right)+\frac{1}{1+c}\ln \left(c\right)\ln \left(1+\frac{1}{c}\right)$$
$$+\frac{1}{1+c}\underbrace{\int _0^{\frac{1}{c}}\frac{\ln \left(1+t\right)}{t}\:dt}_{t=-t}-\frac{1}{1+c}\text{Li}_2\left(-c\right)$$
$$=\frac{2}{1+c}\zeta \left(2\right)-\frac{1}{1+c}\text{Li}_2\left(-\frac{1}{c}\right)-\frac{1}{1+c}\text{Li}_2\left(-c\right)$$
$$=\frac{2}{1+c}\zeta \left(2\right)-\frac{1}{1+c}\left(\text{Li}_2\left(-\frac{1}{c}\right)+\text{Li}_2\left(-c\right)\right)$$
$$=\frac{2}{1+c}\zeta \left(2\right)-\frac{1}{1+c}\left(-\zeta \left(2\right)-\frac{1}{2}\ln ^2\left(c\right)\right)$$
Thus
$$\int_0^{\infty}\frac{\ln\left(x\right)}{\left(c+x\right)\left(x-1\right)}\:dx=\frac{3}{1+c}\zeta \left(2\right)+\frac{1}{2\left(1+c\right)}\ln ^2\left(c\right)$$
