int frac{tan ^{-1} x}{x^3} d x= | vedclass

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(C) Let $I = \int \frac{	an ^{-1} x}{x^3} d x$. Using integration by parts,let $u = 	an ^{-1} x$ and $dv = x^{-3} dx$. Then $du = \frac{1}{1+x^2} dx

Vedclass · Accepted Answer

$\frac{-1}{2 x}-\left(\frac{1}{2}+\frac{1}{2 x^2}ight) 	an ^{-1} x+C$

Vedclass · Answer

(C) Let $I = \int \frac{	an ^{-1} x}{x^3} d x$. Using integration by parts,let $u = 	an ^{-1} x$ and $dv = x^{-3} dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$. $I = u v - \int v du = 	an ^{-1} x \left(-\frac{1}{2x^2}ight) - \int \left(-\frac{1}{2x^2}ight) \frac{1}{1+x^2} dx$. $I = -\frac{	an ^{-1} x}{2x^2} + \frac{1}{2} \int \frac{1}{x^2(1+x^2)} dx$. Using partial fractions,$\frac{1}{x^2(1+x^2)} = \frac{1}{x^2} - \frac{1}{1+x^2}$. $I = -\frac{	an ^{-1} x}{2x^2} + \frac{1}{2} \int \left(\frac{1}{x^2} - \frac{1}{1+x^2}ight) dx$. $I = -\frac{	an ^{-1} x}{2x^2} + \frac{1}{2} \left(-\frac{1}{x} - 	an ^{-1} xight) + C$. $I = -\frac{1}{2x} - \left(\frac{1}{2x^2} + \frac{1}{2}ight) 	an ^{-1} x + C$.

$\int \frac{\tan ^{-1} x}{x^3} d x=$

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