I found here a "great theorem" which states that: $$\int_a^b \frac{\ln(c+dx)}{P(x)}dx =\frac{\ln((ad+c)(bd+c))}{2}\int_a^b\frac{dx}{P(x)}$$ To be fair I dont know how to prove this, but I am pretty sure that we should work by symmetry with a substitution of the form $\frac{mx+n}{sx+p}$, then sum the add the result with the initial integral, because an easier case the well known Putnam integral $\int_0^1 \frac{\ln(1+x)}{1+x^2}dx$ can be dealt with the substitution $\frac{1-x}{1+x}$ which produces $\int_0^1 \frac{\ln 2 -\ln(1+x)}{1+x^2}dx$. So in our case, after finding the magic substitution we will have: $$\int_a^b \frac{\ln(c+dx)}{P(x)}dx=\int_a^b \frac{\ln((ad+c)(bd+c)) - \ln(c+dx)}{P(x)}dx$$ Unfortunately I dont know what $P(x)$ is, but it's hard to believe that it can be any polynomial of the form $x^2+sx+p$. Anyway I am not looking for a full proof, only for a way to "build up" the desired substitution. Login To add answer/comment