Question

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

Answer 1

I would set : `6-x=t^2`
so: `x=6-t^2`
and `dx=-2tdt`
Substituting:
`int(6-t^2)/t(-2t)dt=int(-12+2t^2)dt=-12t+2t^3/3+c`
Going back to `x`:
`=-12sqrt(6-x)+2/3(6-x)sqrt(6-x)+c`