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}

Continuing from yours:
$$I=\int\frac{1}{\sqrt{(x-a)(x-b)}}dx$$
now if we look at this:
$$(x-a)(x-b)=x^2-(a+b)x+ab=\left(x-\frac{a+b}{2}\right)^2-\frac{a^2+b^2}{2}$$
and so we can rewrite our integral as:
$$I=\int\frac{1}{\sqrt{(x-\frac{a+b}{2})^2-\frac{a^2+b^2}{2}}}dx$$
now let $u=x-\frac{a+b}{2}$ so $dx=du$
and the integral becomes:
$$I=\int\frac{1}{\sqrt{u^2-\frac{a^2+b^2}{2}}}du$$
for ease at the moment I am going to let $c^2=\frac{a^2+b^2}{2}$ making our integral:
$$I=\int\frac{1}{\sqrt{u^2-c^2}}du=\frac{1}{c}\int\frac{1}{\sqrt{\left(\frac{u}{c}\right)^2-1}}du$$
now if we let $u=c\cosh(v)$ we get $du=c\sinh(v)dv$ and our integral becomes:
$$I=\frac{1}{c}\int\frac{c\sinh(v)}{\sqrt{\cosh^2(v)-1}}dv=\int1dv$$
$$=v+C=\text{arcosh}\left(\frac{u}{c}\right)+C=\text{arcosh}\left(\frac{x-\frac{a+b}{2}}{c}\right)+C=\text{arcosh}\left(\frac{2x-(a+b)}{a^2+b^2}\right)+C$$
