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Q.
$\int \frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x (x >0)$
Integrals
Solution:
note that $\sec ^{-1} \sqrt{1+x^2}=\tan ^{-1} x$
$\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2 \tan ^{-1} x $ for $x >0$
$I=\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left(\left(\tan ^{-1} x\right)^2+2 \tan ^{-1} x\right) d x $ put $\tan ^{-1} x=t$
$=\int e ^{ t }\left( t ^2+2 t \right) dt = e ^{ t } \cdot t ^2= e ^{\tan ^{-1} x }\left(\tan ^{-1} x \right)^2+ C$