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?

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
$$\int dx \: \frac{\sqrt{\frac{x+1}{x-2}}}{x-2} = -2 (x-2)^{-1/2} (x+1)^{1/2} + \log{\left[\frac{2}{3}\left(x-\frac{1}{2}\right)+\sqrt{\frac{4}{9}\left(x-\frac{1}{2}\right)^2-1}\right]}+C$$
