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

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

Vedclass · Accepted Answer

$(	an ^{-1} x)^2 e^{	an ^{-1} x}+c$,where $c$ is a constant of integration.

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(B) Let $I = \int \frac{e^{	an ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}ight)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight)ight] d x$. Put $x = 	an t$,so $dx = \sec^2 t \, dt$. Since $x > 0$,$t = 	an^{-1} x \in (0, \pi/2)$. Then $\sec^{-1} \sqrt{1+x^2} = \sec^{-1}(\sec t) = t$ and $\cos^{-1}(\frac{1-x^2}{1+x^2}) = \cos^{-1}(\cos 2t) = 2t$. Substituting these into the integral: $I = \int \frac{e^t}{1+	an^2 t} [t^2 + 2t] \sec^2 t \, dt = \int e^t (t^2 + 2t) \, dt$. Using the formula $\int e^t (f(t) + f'(t)) \, dt = e^t f(t) + c$,where $f(t) = t^2$ and $f'(t) = 2t$: $I = e^t \cdot t^2 + c = e^{	an^{-1} x} (	an^{-1} x)^2 + c$.

$\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$,where $x>0$ is

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