#160;Let 1 x^2=t #160; #160; 2xdx=dt #160; #160; #160; #160; #160; #160; #160; #160; #160; xdx= dt2 ∫ ...

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Integration by Parts

∫ xlog 1-x2dx...

Question

$\int x log (1 - x^{2}) d x =$

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Solution

Let $1 - x^{2} = t$
$- 2 x d x = d t$ $x d x = \frac{- d t}{2}$
$\int x log (1 - x^{2}) d x = \frac{- 1}{2} \int log t d t$
$= \frac{- 1}{2} (t log t - \int 1 d t)$
$= \frac{- 1}{2} t (log t - 1) + c$
$= \frac{- 1}{2} (1 - x^{2}) (log (1 - x^{2}) - 1) + c$

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