Mathematics - Indefinite Integration Question with Solution | TestHub

$f(x)=\int \frac{x^{2}dx}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$ Let $x=\tan \theta$ $\Rightarrow dx={\sec }^2\theta d\theta =\left(1+x^2\right)d\theta$ $f(x)=\int \frac{x^{2}dx}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$ $=\int \frac{{\tan }^2\theta {\sec }^2\theta d\theta }{{\sec }^2\theta (1+\sec \theta )}=\int \frac{{\tan }^2\theta d\theta }{1+\sec \theta }$ $=\int \frac{{\sin }^2\theta d\theta }{\cos \theta (1+\cos \theta )}$ $=\int \frac{1-{\cos }^2\theta d\theta }{\cos \theta (1+\cos \theta )}$ $=\int \frac{(1-\cos \theta )d\theta }{\cos \theta }$ $=\int \sec \theta d\theta -\int d\theta$ $=\log \left(x+\sqrt{1+x^2}\right)-{\tan }^{-1}x+C$ $\therefore f(0)=\log (0+\sqrt{1+0})-{\tan }^{-1}(0)+C$ $0=\log 1-0+C$ $\Rightarrow C=0$ $\therefore f(1)=\log \left(1+\sqrt{1+1^2}\right)-{\tan }^{-1}(1)$ $=\log (1+\sqrt{2})-\frac{\pi }{4}$