How can I integrate $\int\frac{1}{\sqrt{(x-a)(x-b)}}dx$ How can I solve the differential equation:
$$
\frac{dy}{dx}=\frac{1}{\sqrt{(x-a)(x-b)}}
$$
where $a < b$ and $a>0$.
So this question is basically asking the integral of the above equation over $x$.
Thanks in advance.
 A: setting $$\sqrt{(x-a)(x-b)}=x+t$$ after squaring and solving for $x$ we get
$$x=\frac{ab-t^2}{2t+a+b}$$ and we get
$$x+t=\frac{(b+t)(a+t)}{2t+a+b}$$ and
$$dx=-\frac{2(b+t)(a+t)}{(2t+a+b)^2}dt$$
A: There are really many substitutions that works for this integral. My suggestion is:
$$
\sqrt{x-b}=u \quad \rightarrow \quad dx=2udu
$$
and, since:
$$
x-b=u^2 \quad \rightarrow \quad \sqrt{u-a}=\sqrt{u^2+b-a}
$$
the integral becomes
$$
2\int\frac{1}{\sqrt {u^2+c^2}} du
$$
(where $c=\sqrt{b-a}$). And this a well known integral.
A: At the risk of stating the obvious, you compute the integral using the methods you learned in calculus class for integrals of this form.
Maybe you have a mental block, because you haven't realized $(x-a)(x-b)$ is a quadratic polynomial in $x$. Maybe writing it as $x^2 - (a+b)x + ab$ would jog your memory. (and possibly plugging in values for $a,b$ to consider a special case)
A: Just as Hurkyl answered, start with $$(x-a)(x-b)=x^2 - (a+b)x + ab=\left( x-\frac{a+b}2\right)^2-\left(\frac{a-b}2 \right)^2$$ Now define $$x-\frac{a+b}2=\frac{a-b}2 t\implies x=\frac{a+b}2+\frac{a-b}2 t\implies dx=\frac{a-b}2 dt$$ This makes $$\frac{dx}{\sqrt{(x-a)(x-b)}}=\frac{dt}{\sqrt{t^2-1}}$$ which is well known.
A: Assume $x>b>a$ and substitute $ t =\sqrt{x-a} + \sqrt{x-b}$. Then
$$\frac{1}{\sqrt{(x-a)(x-b)}}dx=\frac2tdt
$$
and
\begin{align}\int \frac{1}{\sqrt{(x-a)(x-b)}}dx
=\int\frac2tdt=2\ln( \sqrt{x-a} + \sqrt{x-b})+C
\end{align}
