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}

Alternatively, you can use an Euler substitution to rationalize the integrand.

Option 1 Change variable to $t$, where $$\sqrt{(x - a) (x - b)} = x + t.$$
rearranging gives $$x = \frac{ab - t^2}{2 t + (a + b)},$$
and substituting gives $$\int \frac{dx}{\sqrt{(x - a) (x - b)}} = \int \frac{dt}{t + \frac{1}{2}(a + b)} .$$

Option 2 Change variable to $u$, where $$\sqrt{(x - a) (x - b)} = (x - a) u.$$ Rearranging gives
$$x = \frac{a u^2 - b}{u^2 - 1},$$
and substituting gives $$-2 \int \frac{du}{u^2 - 1} .$$
