Mathematics

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

##### SOLUTION
Let $1-x^{2}=t$
$-2xdx=dt$                   $xdx=\dfrac{-dt}{2}$
$\int x\log (1-x^{2})dx= \dfrac{-1}{2}\int\log tdt$
$=\dfrac{-1}{2}(t\log t-\int1dt)$
$=\dfrac{-1}{2}t(\log t-1) +c$
$=\dfrac{-1}{2}(1-x^{2})(\log (1-x^{2})-1)+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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