Integral of irrational function $$
\int \frac{\sqrt{\frac{x+1}{x-2}}}{x-2}dx
$$
I tried:
$$
t =x-2
$$
$$
dt = dx
$$
but it didn't work.
Do you have any other ideas?
 A: Hint: $$\frac{x+1}{x-2} = 1 + \frac{3}{x-2}$$
The substitution can be done then as $u=\sqrt{\frac{x+1}{x-2}}$, which can be written as $x=\frac{3}{u^2-1}+2$. The resulting integral is a little messy, but avoids trigonometric substitutions.
You end up with the simple:
$$\int \left(-2 +\frac{1}{u+1}-\frac{1}{u-1}\right)du$$
Yielding: $$-2u + \log{\frac{u+1}{u-1}} + C$$
You can, with some work, see that $\frac{u+1}{u-1}$ is actually the same value as inside the log in the other answer above, by writing it as $\frac{(u+1)^2}{u^2-1}$ and replacing values. In paricular, $u^2=1+\frac{3}{x-2}$. You get $$\frac{u+1}{u-1} = \frac{2x-1 + 2\sqrt{(x+1)(x-2)}}{3}$$
Note
You could first do the substitution $v=\frac{1}{x-2}$ which is the substitution most obvious from my hint, getting $x=2+\frac{1}{v}$. You end up with:
$$-\int \frac{\sqrt{3v+1}}{v}dv$$
But from there, you really need to substitute $w=\sqrt{3v+1}$ to get an integral, and that $w$ is the same as our original $u$. Still, might be easier to do it in two steps.
A: Let $y=x-2$ and integrate by parts to get
$$\int dx \: \frac{\sqrt{\frac{x+1}{x-2}}}{x-2} = -2 (x-2)^{-1/2} (x+1)^{1/2} + \int \frac{dy}{\sqrt{y (y+3)}}$$
In the second integral, complete the square in the denominator to get
$$\int \frac{dy}{\sqrt{y (y+3)}} = \int \frac{dy}{\sqrt{(y+3/2)^2-9/4}}$$
This integral may be solved using a substitution $y+3/2=3/2 \sec{\theta}$, $dy = 3/2 \sec{\theta} \tan{\theta}$.  Using the fact that
$$\int d\theta \sec{\theta} = \log{(\sec{\theta}+\tan{\theta})}+C$$
we may evaluate the integral exactly.  I leave the intervening steps to the reader; I get
$$\int dx \: \frac{\sqrt{\frac{x+1}{x-2}}}{x-2} = -2 (x-2)^{-1/2} (x+1)^{1/2} + \log{\left[\frac{2}{3}\left(x-\frac{1}{2}\right)+\sqrt{\frac{4}{9}\left(x-\frac{1}{2}\right)^2-1}\right]}+C$$
