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

Let \(\mathrm{I}=\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x\)
\(\text { Put } x=\tan t \)
\( \therefore \mathrm{d} x=\sec ^2 \mathrm{t} d \mathrm{dt}\)
\(\therefore I=\int \frac{e^{\tan ^{-1}(\tan t)}}{1+\tan ^2 t}[\left(\sec ^{-1} \sqrt{1+\tan ^2 t}\right)^2+\) \(\cos ^{-1}\left(\frac{1-\tan ^2 t}{1+\tan ^2 t}\right)] \)
\( \sec ^2 t d t\)
\(=\int \frac{e^t}{\sec ^2 t}\left[\left(\sec ^{-1}(\sec t)\right)^2+\cos ^{-1}(\cos 2 t)\right] \sec ^2 t d t \)
\( =\int e^t\left[t^2+2 t\right] d t \)
\( =e^t \cdot t^2+c\)
\(=\mathrm{t}^2 \cdot \mathrm{e}^{\mathrm{t}}+\mathrm{c}^{\left[\left[\mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\right.} \)
\( =\left(\tan ^{-1} x\right)^2 \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}\)