How find this integral $I=\int_{-1}^{1}\frac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}$ 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 result is nice

 A: Notice for $|t| < 1$, $\frac{1}{\sqrt{1 - 2xt + t^2}}$ is the generating function for the Legendre polynomials:
$$\frac{1}{\sqrt{1 - 2xt + t^2}} = \sum_{n=0}^\infty P_n(x) t^n\tag{*1}$$
It is known that for $x \in [-1,1]$,  $|P_n(x)| \le P_n(1) = 1$. This means as long as $t$ is fixed and $|t| < 1$, the absolute values of the $n^{th}$ term is bounded by a geometric series $|t|^n$. As a result, the convergence in $(*1)$ is absolute and uniform for $x$ over $[-1,1]$. 
Recall the orthogonality relation for Legendre polynomials:
$$\int_{-1}^1 P_n(x) P_m(x) dx = 
\begin{cases} \frac{2}{2n+1}, & n = m\\ 0, & n \ne m\end{cases}
$$
We find
$$\begin{align}
&\int_{-1}^1 \frac{dx}{\sqrt{1-2ax+a^2}\sqrt{1-2bx+b^2}}\\
=& \int_{-1}^1 \left(\sum_{n=0}^\infty P_n(x) a^n\right)\left(\sum_{m=0}^\infty P_m(x) b^m\right) dx\\
=& \sum_{n=0}^\infty \sum_{m=0}^\infty a^n b^m \int_{-1}^1 P_n(x) P_m(x) dx\\
=& \sum_{n=0}^\infty \frac{2}{2n+1} (ab)^n\\
=& \sum_{k=0}^\infty \frac{1}{(k+1)}\left( (\sqrt{ab})^k + (-\sqrt{ab})^k \right)\\
=& - \frac{ \log(1 - \sqrt{ab})}{\sqrt{ab}} + \frac{ \log(1 + \sqrt{ab})}{\sqrt{ab}}\\
=& \frac{1}{\sqrt{ab}}\log\left(\frac{1+\sqrt{ab}}{1-\sqrt{ab}}\right)
\end{align}
$$
A: 
$$$$
  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.

A: \begin{eqnarray}
& &\int_{-1}^{1}\dfrac{1}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}dx\\
&=&-\frac{1}{\sqrt{ab}}\tanh^{-1}\sqrt{\frac{{b(a^2+1)-2abx}}{{a(b^2+1)-2abx}}}\bigg|_{-1}^1\\
&= &\frac{1}{\sqrt{ab}}\tanh^{-1}\dfrac{2\sqrt{ab}}{1+ab}
=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}
\end{eqnarray}
