TITLE: How find this integral $I=\int_{-1}^{1}\frac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}$
QUESTION [10 upvotes]: show this integral
$$I=\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}$$
where $0<a,b<1$
my idea: let
\begin{align*}&(-2ax+a^2+1)(-2bx+b^2+1)=4abx^2-2(a+b+a^2b+b^2a)x+(a^2+1)(b^2+1)\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-\dfrac{4ab(a+b+a^2b+b^2a)^2}{4ab}\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-(a+b+a^2b+b^2a)^2\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a+b)^2+(ab-1)^2+(a+b)^2(1+ab)^2
\end{align*}
so I think this idea is not good, maybe this have good methods,because this reslut is nice

REPLY [9 votes]: $$$$
  Let us start to calculate it.
  \begin{eqnarray}
I&=&\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}\\
&=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{\frac{a^2+1}{2a}-x}\sqrt{\frac{b^2+1}{2b}-x}}\\
&=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{m-x}\sqrt{n-x}}\\ 
&=&\frac{1}{\sqrt{ab}}\int_{-1}^{1}\dfrac{-d(\sqrt{n-x})}{\sqrt{(\sqrt{n-x})^2+m-n}}\\
&=&-\frac{1}{\sqrt{ab}}\ln(\sqrt{n-x}+\sqrt{m-x}\sqrt{n-x})|_{-1}^{1}\\
&=&\frac{1}{\sqrt{ab}}\frac{ln(\sqrt{n+1}+\sqrt{m+1}\sqrt{n+1})}{ln(\sqrt{n-1}+\sqrt{m-1}\sqrt{n-1})}\\
&=&\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}
\end{eqnarray} 
  Where $m=\frac{a^2+1}{2a} \\
n=\frac{b^2+1}{2b}$
So we can get it.