int frac{e^{tan ^{-1} x}}{1+x^2}left[left(sec | vedclass

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(A) Let $I = \int \frac{e^{	an ^{-1} x}}{1+x^2} \left[ (\sec ^{-1} \sqrt{1+x^2})^2 + \cos ^{-1} \left( \frac{1-x^2}{1+x^2} ight) ight] dx$. Since

Vedclass · Accepted Answer

$e^{	an ^{-1} x}(	an ^{-1} x)^2+C$

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(A) Let $I = \int \frac{e^{	an ^{-1} x}}{1+x^2} \left[ (\sec ^{-1} \sqrt{1+x^2})^2 + \cos ^{-1} \left( \frac{1-x^2}{1+x^2} ight) ight] dx$. Since $\sec ^{-1} \sqrt{1+x^2} = 	an ^{-1} x$ and $\cos ^{-1} \left( \frac{1-x^2}{1+x^2} ight) = 2 	an ^{-1} x$,the integral becomes: $I = \int \frac{e^{	an ^{-1} x}}{1+x^2} [(	an ^{-1} x)^2 + 2 	an ^{-1} x] dx$. Let $t = 	an ^{-1} x$,then $dt = \frac{1}{1+x^2} dx$. $I = \int e^t (t^2 + 2t) dt = \int (t^2 e^t + 2t e^t) dt$. Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = t^2$ and $f'(t) = 2t$: $I = e^t t^2 + C = e^{	an ^{-1} x} (	an ^{-1} x)^2 + C$.

$\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x=$

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