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

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

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Answer 1 · 2015-03-01T22:00:24

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 $x$ :
$=-12sqrt(6-x)+2/3(6-x)sqrt(6-x)+c$

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

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