How do I evaluate the integral $\int \frac{x}{\sqrt{6 - x}} d x$ $intx/(sqrt(6-x)) dx$ ?

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How do I evaluate the integral $\int \frac{x}{\sqrt{6 - x}} d x$ $intx/(sqrt(6-x)) dx$ ?

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Answer 1 · 2015-03-01T22:00:24

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

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How do I evaluate the integral $\int \frac{x}{\sqrt{6 - x}} d x$ $intx/(sqrt(6-x)) dx$ ?

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