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#