# How do I evaluate the integral intx/(sqrt(6-x)) dx?

I would set : $6 - x = {t}^{2}$
so: $x = 6 - {t}^{2}$
and $\mathrm{dx} = - 2 t \mathrm{dt}$
$\int \frac{6 - {t}^{2}}{t} \left(- 2 t\right) \mathrm{dt} = \int \left(- 12 + 2 {t}^{2}\right) \mathrm{dt} = - 12 t + 2 {t}^{3} / 3 + c$
Going back to $x$:
$= - 12 \sqrt{6 - x} + \frac{2}{3} \left(6 - x\right) \sqrt{6 - x} + c$