I wish to calculate
$$\int \frac{x}{\sqrt{x^{2}-36}} \, {\rm d} x$$
I use a substitution $x=6\sec\theta$ and get $\int \csc\theta d\theta$ as intermediate.. Then, after some calculation
$$-\ln\left|\frac{x}{\sqrt{x^2-36}}+\frac{6}{x^2-36}\right|$$
Is it correct?
Go easy on it. Take
$\displaystyle x^2 - 36 = y \ \implies 2x \ \mathrm {d}x = \mathrm {d}y$
$$\displaystyle \int \frac {x}{\sqrt {x^2 -36}} \ \mathrm {d}x = \frac {1}{2}\int y^{-1/2} \ \mathrm {d}y = \sqrt {y} + c .$$
Substitute the value of $y$ and you get :
$$\displaystyle \int \frac {x}{\sqrt {x^2 -36}} \ \mathrm {d}x = \sqrt {x^2 -36} + c .$$
But as mentioned by @integration brainstorm : A typo in the Question, using same substitution :
$$\displaystyle \int \frac {x}{{x^2 -36}} \ \mathrm {d}x = \frac {1}{2}\int y^{-1} \ \mathrm {d}y = \frac {\ln {y}}{2} + c . $$
Thus $\displaystyle \int \frac {x}{{x^2 -36}} \ \mathrm {d}x = \frac {\ln (x^2 -36)}{2} + c . $