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}
 A: 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 $$-2 \int \frac{du}{u^2 - 1} .$$
A: 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:
$$I=\frac{1}{c}\int\frac{c\sinh(v)}{\sqrt{\cosh^2(v)-1}}dv=\int1dv$$
$$=v+C=\text{arcosh}\left(\frac{u}{c}\right)+C=\text{arcosh}\left(\frac{x-\frac{a+b}{2}}{c}\right)+C=\text{arcosh}\left(\frac{2x-(a+b)}{a^2+b^2}\right)+C$$
A: Find $p$ and $q$ such that $(x-a)(x-b) = (x-p)^2 \pm q^2$. Then substitute $u = x-p$, and you get a standard integral that I learned in high school. See #32 in this table. 
Or, alternatively, just use formula #39 from that same table.
A: 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,
$$J=2\int{\frac{du}{4-{{u}^{2}}}}$$
A: Proceeding further from where you left 
$$
\int \dfrac {2\,}{\sqrt {a-b+t^2}}\ dt
$$
$$
\int \dfrac {2}{\sqrt {(t)^2 + (\sqrt{a-b}) ^2}}\ dt
$$
Using
$$
\int \frac{1}{\sqrt{x^2 + a ^ 2}} = \log({x+\sqrt{x^2+a^2}})
$$
It becomes 
$$
2 \cdot {\log (t+\sqrt{t^2 + (\sqrt{a-b}) ^2})}
$$
Then substitute back the value of t .
And you get the required result .
