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
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