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I came across the following integral
$$\int_{0}^1 \dfrac{\log(x)}{(x-a)(x-b)}dx=\dfrac{1}{a-b}\left(\text{Li}_2\left(\dfrac{1}{a}\right)-\text{Li}_2\left(\dfrac{1}{b}\right)\right)$$
but I haven't been able to prove it. I'm trying to use partial fractions/some normal substitutions and tried to write the answer in the required dilogarithm form, but it seems a bit tricky. Is there an easy proof of this identity?
As you did $$\frac{1}{(x-a) (x-b)}=\frac 1{a-b}\left(\frac 1{x-a}-\frac 1{x-b} \right)$$ So, we face two integrals $$I=\int \frac {\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{x}{c}\right)+\log (x) \log \left(1-\frac{x}{c}\right)$$ (one integration by parts). $$J=\int_0^1 \frac {\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{1}{c}\right)$$
$$\int_0^1\frac{\ln x}{a-x}\ dx=\frac1a\int_0^1\frac{\ln x}{1-x/a}\ dx$$
$$=\frac1a\sum_{n=1}^\infty\frac1{a^{n-1}}\int_0^1 x^{n-1}\ln x\ dx=\sum_{n=1}^\infty\frac{1}{a^n}\left(-\frac{1}{n^2}\right)$$ $$=-\sum_{n=1}^\infty\frac{1}{a^n n^2}=-\operatorname{Li}_2\left(\frac{1}{a}\right)$$
