How do you find the integral of $(x/(sqrt(6-x))) dx$ ?

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Answer 1 · 2015-02-18T15:28:59

I would set: $6-x=t^2$

So that: $x=6-t^2$ and $dx=-2tdt$

Substituting for $x$ the integral becomes:

$int(6-t^2)/t(-2t)dt=$

$=-2int(6-t^2)dt=-2[6t-t^3/3]+c=$

Going back to $x$ :

$=-12sqrt(6-t)+2/3(6-t)^(3/2)+c$

How do you find the integral of (x/(sqrt(6-x))) dx(x/(sqrt(6-x))) dx?