TITLE: Integral of the form $\int_a^b \frac{\ln(c+dx)}{P(x)}dx$
QUESTION [8 upvotes]: I found here a "great theorem" which states that: $$\int_a^b \frac{\ln(c+dx)}{P(x)}dx =\frac{\ln((ad+c)(bd+c))}{2}\int_a^b\frac{dx}{P(x)}$$
I don't 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 add the result with the initial integral.
An easier case which shows this idea is the well-known integral $\int_0^1 \frac{\ln(1+x)}{1+x^2}dx$ which can be dealt with the substitution $\frac{1-x}{1+x}$ which produces $\int_0^1 \frac{\ln 2 -\ln(1+x)}{1+x^2}dx$ and adding this with the initial integral simplifies the logarithm. 
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 $x^2+sx+p$. I would appreciate some help to prove this "great theorem".

REPLY [8 votes]: I think I finally found it! I will denote the differential with $\textbf{dx}$ to not get confused.
$$I=\int_a^b \frac{\ln(c+dx)}{P(x)}\textbf{dx}$$
Using the following substitution:
$$x=\frac{a(c+bd)+c(b-t)}{c+dt}\Rightarrow \textbf{dx}=-\frac{(c+ad)(c+bd)}{(c+dt)^2}\textbf{dt}$$
Also $\,\displaystyle{t=\frac{a(c+bd)+c(b-x)}{c+dx}}\, $ thus if we plug in the bounds we get:$$x=a\rightarrow t=\frac{b(c+ad)}{c+ad}=b$$
$$x=b\rightarrow t=\frac{a(c+bd)}{(c+bd)}=a$$
$$I=\int_a^b \frac{\ln\left(c+d\left(\frac{a(c+bd)+c(b-t)}{c+dt}\right)\right)}{P\left(\frac{a(c+bd)+c(b-t)}{c+dt}\right)}\frac{(c+ad)(c+bd)}{(c+dt)^2}\textbf{dt}$$$$\overset{t=x}=\int_a^b \frac{\ln((c+ad)(c+bd))-\ln(c+dx)}{Q(x)}\mathbf{dx}$$
Of course this holds only if $\displaystyle{Q(x)=P(x)=P\left(\frac{a(c+bd)+c(b-x)}{c+dx}\right)}\frac{(c+dx)^2}{(c+ad)(c+bd)}\,$ but if this happens then if we have that: $$I=\frac{\ln((c+ad)(c+bd)}{2}\int_a^b\frac{dx}{P(x)}$$