int frac{x}{left(1-x^2right) sqrt{2-x^2}} d x | vedclass

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(C) Let $I = \int \frac{x}{(1-x^2) \sqrt{2-x^2}} dx$. Substitute $t = \sqrt{2-x^2}$. Then $t^2 = 2-x^2$,which implies $x^2 = 2-t^2$. Differentiating b

Vedclass · Accepted Answer

$\frac{1}{2} \log \left|\frac{1+\sqrt{2-x^2}}{1-\sqrt{2-x^2}}\right|+c$

Vedclass · Answer

(C) Let $I = \int \frac{x}{(1-x^2) \sqrt{2-x^2}} dx$. Substitute $t = \sqrt{2-x^2}$. Then $t^2 = 2-x^2$,which implies $x^2 = 2-t^2$. Differentiating both sides,$2t dt = -2x dx$,so $x dx = -t dt$. Substituting these into the integral: $I = \int \frac{-t dt}{(1-(2-t^2)) t} = \int \frac{-dt}{t^2-1} = \int \frac{dt}{1-t^2}$. Using the standard integral formula $\int \frac{dt}{a^2-t^2} = \frac{1}{2a} \log \left| \frac{a+t}{a-t} \right| + c$ with $a=1$: $I = \frac{1}{2} \log \left| \frac{1+t}{1-t} \right| + c$. Substituting $t = \sqrt{2-x^2}$ back: $I = \frac{1}{2} \log \left| \frac{1+\sqrt{2-x^2}}{1-\sqrt{2-x^2}} \right| + c$. Thus,the correct option is $C$.

$\int \frac{x}{\left(1-x^2\right) \sqrt{2-x^2}} d x=$

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