Is there any way to integrate the following?:
$\displaystyle \int \frac{dx}{\sqrt{a^{-1}-x^{-1}}}$
(Where "a" is a constant.)
Any help would be appreciated.
Fist we would need to simplfy to get
$\displaystyle \frac{1}{\sqrt{\frac{1}{a}-\frac{1}{x}}}=\frac{1}{\sqrt{\frac{x-a}{ax}}}=\frac{\sqrt{ax}}{\sqrt{x-a}}$
So we have the integral
$\displaystyle \sqrt{a}\int \frac{\sqrt{x}}{\sqrt{x-a}}dx$
Now we let $\displaystyle x=a\sec^2(t) \implies dx=2a\sec(t)[\sec(t)\tan(t)]dt$
$\displaystyle \sqrt{a}\int \frac{\sqrt{x}}{x-a}dx=\sqrt{a}\int \frac{\sqrt{a}\sec(t)}{\sqrt{a\sec^2(t)-a}}(2a\sec^2(t)\tan(t)dt)=$
$\displaystyle 2a^{\frac{3}{2}}\int \sec^{3}(t)dt$
Here is how you integrate $\displaystyle \sec^{3}(t)$
Integral of secant cubed - Wikipedia, the free encyclopedia
So we get
$\displaystyle 2a^{3/2}\left[\frac{1}{2}\sec(t)\tan(t)+\frac{1}{2}\ln|\sec(t)+\ tan(t)| \right]$
From here we just need to back substitue
$\displaystyle x=a\sec^2(t) \iff \frac{x}{a}=\sec^2(t) \iff \frac{x}{a}=\tan^2(t)+1$
$\displaystyle \tan^2(t)=\frac{x-a}{a} \iff \tan(t)=\sqrt{\frac{x-a}{a}}$ and of course
$\displaystyle \sec(t) =\sqrt{\frac{x}{a}}$
I will leave the rest to you