<s>[INST] Integrate $\int \frac {dx}{\sqrt {(x-a)(x-b)}}$ where $b>a$ Integrate: $\displaystyle\int \dfrac {dx}{\sqrt { (x-a)(x-b)}}$ where $b>a$
My Attempt:
$$\int \dfrac {dx}{\sqrt {(x-a)(x-b)}}$$
Put $x-a=t^2$
$$dx=2t\,dt$$
Now,
\begin{align}
&=\int \dfrac {2t\,dt}{\sqrt {t^2(a+t^2-b)}}\\
&=\int \dfrac {2\,dt}{\sqrt {a-b+t^2}}
\end{align}
 [/INST] Set 
$$J=\int{\frac{dx}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}}$$
Use the following substitution: 
$$u=\frac{2x-\left( a+b \right)}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}$$
once get:
$$du=\left( -\frac{{{\left( 2x-\left( a+b \right) \right)}^{2}}}{2{{\left( \sqrt{{{x}^{2}}-\left( a+b \right)x+ab} \right)}^{3}}}+\frac{2}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}} \right)dx$$
now use 
$$\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}=\frac{2x-\left( a+b \right)}{u}$$
and after some algebraic simplifications you should have:
$$dx=\frac{2\left( 2x-\left( a+b \right) \right)}{4u-{{u}^{3}}}du$$
Finally,
$$J=2\int{\frac{du}{4-{{u}^{2}}}}$$
</s>