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