$\int \frac{\mathrm{e}^{ an ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2} ight)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2} ight) ight] \mathrm{d} x$, where $x\gt0$ is

Question

$\int \frac{\mathrm{e}^{	an ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}ight)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight)ight] \mathrm{d} x$, where $x\gt0$ is

Aicharya · Accepted Answer

Let $\mathrm{I}=\int \frac{\mathrm{e}^{	an ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}ight)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight)ight] \mathrm{d} x$ $\begin{array}{ll} & 	ext { Put } x=	an t \ 	herefore \quad & \mathrm{~d} x=\sec ^2 \mathrm{t} d \mathrm{dt}\end{array}$ $\begin{array}{r}	herefore \quad I=\int \frac{e^{	an ^{-1}(	an t)}}{1+	an ^2 t}\left[\left(\sec ^{-1} \sqrt{1+	an ^2 t}ight)^2+\cos ^{-1}\left(\frac{1-	an ^2 t}{1+	an ^2 t}ight)ight] \ \sec ^2 t d t\end{array}$ $\begin{aligned} & =\int \frac{e^t}{\sec ^2 t}\left[\left(\sec ^{-1}(\sec t)ight)^2+\cos ^{-1}(\cos 2 t)ight] \sec ^2 t d t \ & =\int e^t\left[t^2+2 tight] d t \ & =e^t \cdot t^2+c\end{aligned}$ $\begin{aligned} & =\mathrm{t}^2 \cdot \mathrm{e}^{\mathrm{t}}+\mathrm{c}^{\left[\left[\mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)ight] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}ight]ight.} \ & =\left(	an ^{-1} xight)^2 \mathrm{e}^{	an ^{-1} x}+\mathrm{c}\end{aligned}$

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

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