The value of int frac{e^{tan ^{-1}(x)}}{1+x^2 | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{e^{	an ^{-1} x}}{1+x^2} \left[\left(\sec ^{-1} \sqrt{1+x^2}ight)^2 + \cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight)ight] d x$ for $x > 0$. Substitute $	an ^{-1} x = 	heta$,so $x = 	an 	heta$ and $\frac{1}{1+x^2} d x = d 	heta$. Since $x > 0$,$	heta \in (0, \pi/2)$. Note that $\sec ^{-1} \sqrt{1+x^2} = \sec ^{-1} \sqrt{1+	an^2 	heta} = \sec ^{-1} \sec 	heta = 	heta$. Also,$\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight) = \cos ^{-1}(\cos 2	heta) = 2	heta$ (since $2	heta \in (0, \pi)$). Substituting these into the integral: $I = \int e^{	heta} [	heta^2 + 2	heta] d 	heta$. Using the standard integral formula $\int e^x [f(x) + f'(x)] d x = e^x f(x) + c$,where $f(	heta) = 	heta^2$ and $f'(	heta) = 2	heta$: $I = e^{	heta} 	heta^2 + c$. Substituting back $	heta = 	an ^{-1} x$: $I = e^{	an ^{-1} x} (	an ^{-1} x)^2 + c$.

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

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