How to solve $\int_{0}^{1} \frac{(x-1)\ln x}{(x+1)(x^2+2\cosh \alpha+1)} \,dx$ I have this unpleasant integral that appear in my last exam which i was not able solve

For $\alpha \in \mathbb{R}$. Evaluate $$I=\int_{0}^{1} \frac{(x-1)\ln x}{(x+1)(x^2+2\cosh \alpha+1)} \,dx$$

Anyway my attempt is that I try to do some partial fraction on the integral
$$\frac{(x-1)}{(x+1)(x^2+2\cosh \alpha+1)} = \frac{(x-1)}{(x+1)(x+\cosh \alpha-\sinh \alpha)(x+\cosh \alpha+\sinh \alpha)}$$
$$\frac{(x-1)}{(x+1)(x+\cosh \alpha-\sinh \alpha)(x+\cosh \alpha+\sinh \alpha)}=\frac{a}{x+1}+\frac{b}{x+\cosh \alpha-\sinh \alpha}+\frac{c}{x+\cosh \alpha+\sinh \alpha}$$
And I find that
$$a=\frac{1}{\cosh \alpha-1}, b=c=\frac{1}{2-2\cosh \alpha}$$
That's all I could do. I still don't know what to do next

It is not so unpleasant. To make things easier (at least to me), consider
$$\frac{x-1}{(x+1)(x^2+k)}=\frac{x-1}{(x+1) (x-a) (x-b)}$$ and, as you did, using partial fraction decomposition, this becomes
$$\frac{a-1}{(a+1) (a-b) (x-a)}-\frac{2}{(a+1) (b+1) (x+1)}+\frac{b-1}{(b+1) (b-a)   (x-b)}$$ One integration by parts leads to
$$\int\frac {\log(x)}{x+c} dx=\text{Li}_2\left(-\frac{x}{c}\right)+\log (x) \log \left(1+\frac{x}{c}\right)$$
$$\int_0^1\frac {\log(x)}{x+c} dx=\text{Li}_2\left(-\frac{1}{c}\right)$$ Combining all of the above
$$I=\int_0^1\frac{(x-1)\log(x)}{(x+1) (x-a) (x-b)}=$$
$$\frac{6 (a-1) (b+1) \text{Li}_2\left(\frac{1}{a}\right)-6 (a+1) (b-1)
   \text{Li}_2\left(\frac{1}{b}\right)+\pi ^2 (a-b)}{6 (a+1) (b+1) (a-b)}$$
As you found
$$a=-e^{-\alpha} \qquad \text{and} \qquad b=-e^{\alpha}$$ Replacing leads to
$$I=-\frac{e^{\alpha } \left(6 \text{Li}_2\left(-e^{-\alpha }\right)+6
   \text{Li}_2\left(-e^{\alpha }\right)+\pi ^2\right)}{6 \left(e^{\alpha
   }-1\right)^2}$$ which looks like a gaussian.
