Mathematics - Indefinite Integration Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Let }f(x)=\int \left(\frac{x^2+{\sin }^{2}x}{1+x^2}\right){\sec }^{2}xdx\\ & =\int \frac{x^2{\sec }^{2}x+\frac{{\sin }^{2}x}{{\cos }^{2}x}}{1+x^2}dx\end{array}$$ $$\begin{array}{rl}& =\int \frac{x^2{\sec }^{2}x+{\tan }^{2}x}{1+x^2}dx\\ & =\int \frac{x^2\left(1+{\tan }^{2}x\right)+{\tan }^{2}x}{1+x^2}dx\\ & =\int \frac{x^2+{\tan }^{2}x\left(1+x^2\right)}{1+x^2}dx\\ & =\int \frac{x^2}{1+x^2}dx+\int {\tan }^{2}xdx\\ & =\int \frac{x^2+1-1}{1+x^2}\mathrm{dx}+\int ({\sec }^2\mathrm{x}-1\mathrm{dx}\\ & =\int 1dx-\int \frac{dx}{1+x^2}+\int {\sec }^{2}xdx-\int dx\\ & =-\tan -1x+{\tan }^{-1}x+\mathrm{c}\\ & \text{ Given: }f(0)=0\\ & \Rightarrow f(0)=-{\tan }^{-1}0+{\tan }^2+c\\ & \Rightarrow c=0\\ & \therefore f(x)=-{\tan }^{-1}x+{\tan }^{2}x\end{array}$$ Now, $$f(1)=-{\tan }^{-1}(1)+\tan 1=\tan 1-\frac{\pi }{4}$$