Integral of the form $\int_a^b \frac{\ln(c+dx)}{P(x)}dx$

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".

Solution 1:

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: $$I=\frac{\ln((c+ad)(c+bd)}{2}\int_a^b\frac{dx}{P(x)}$$

Solution 2:

\begin{align}J=\int_a^b \frac{\ln(c+dx)}{P(x)}dx\end{align}

Formally,

1)"Clean up" the logarithm. Perform the change of variable $u=c+dx$,

\begin{align}J=\frac{1}{d}\int_{c+da}^{c+db} \frac{\ln u}{P\left(\frac{u-c}{d}\right)}du\end{align}

2) Change of the bounds of the integral to new ones, m,M such that $m\times M=1$.

Perform the change of variable $v=\frac{1}{\sqrt{(c+db)(c+da)}}u$

\begin{align}J&=\frac{1}{d}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{\ln\left( v\sqrt{(c+db)(c+da)}\right)}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv\\ &=\frac{1}{d}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{\ln v}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv+\\ &\frac{\ln\left((c+db)(c+da)\right)}{2d}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{1}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv \end{align}

In the latter integral perform the change of variable $z=\dfrac{v\sqrt{(c+db)(c+da)}-c}{d}$,

\begin{align}J&=\frac{1}{d}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{\ln v}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv+\frac{\ln\left((c+db)(c+da)\right)}{2}\int_a^b\frac{1}{P(z)}\,dz\end{align}

If for all $v$ real, $v^2 P\left(\frac{\frac{1}{v}\sqrt{(c+db)(c+da)}-c}{d}\right)=P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)$ then ,

\begin{align}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{\ln v}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv=0\end{align}

(Perform the change of variable $w=\dfrac{1}{v}$)

Thus,

\begin{align}J=\frac{\ln\left((c+db)(c+da)\right)}{2}\int_a^b\frac{1}{P(z)}\,dz\end{align}

PS: Don't expect to use this formula with P a polynomial of degree>2 or even equal to 1.