The value of $\int \frac{e^{ an ^{-1}(x)}}{1+x^2}$ $\left[\left(\sec ^{-1} \sqrt{1+x^2} ight)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2} ight) ight] d x$, for $x0$ is

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

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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(x>0) $ Let us take $	an ^{-1} x=	heta \Rightarrow x=	an 	heta$ $ \begin{aligned} & \frac{1}{1+x^2} d x=d 	heta \ & \mathrm{I}=\int e^	heta\left[\left(\sec ^{-1}\left(\sqrt{1+	an ^2 	heta}ight)ight)^2ight. \ & \left.+\cos ^{-1}\left(\frac{1-	an ^2 	heta}{1+	an ^2 	heta}ight)ight] d 	heta \ & =\int e^	heta\left\{\left(\sec ^{-1} \sec 	hetaight)^2+\left(\cos ^{-1} \cos 2 	hetaight)ight\} d 	heta \ & {\left[\because \cos 2 	heta=\frac{1-	an ^2 	heta}{1+	an ^2 	heta}ight]} \ & =\int e^	heta\left\{	heta^2+2 	hetaight\} d 	heta\left[\because \sec ^{-1} \sec 	heta=	hetaight. 	ext { and } \ & \end{aligned} $ $ \begin{aligned} & \cos ^{-1} \cos 2 	heta=2 	heta] \ & {\left[\because \int e^x\left(f(x)+f^{\prime}(x)ight] d x=e^x f(x)+cight] } \end{aligned} $ So, $ \begin{aligned} & I=e^	heta \cdot 	heta^2+c \ & I=e^{	an ^{-1} x} \cdot\left(	an ^{-1} xight)^2+c \end{aligned} $

The value of $\int \frac{e^{\tan ^{-1}(x)}}{1+x^2}$ $\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos…

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