Mathematics - Definite Integration Question with Solution | TestHub

(c) We note that $\frac{d}{dx}(x\tan x+1)=x{\sec }^{2}x+\tan x$ $\therefore$ integrating by parts with $x^2$ as first function, we get $\begin{array}{rl}& I=\int x^2\frac{x{\sec }^{2}x+\tan x}{(x\tan x+1{)}^2}dx\\ & =x^2\left(-\frac{1}{x\tan x+1}\right)-\int 2x\left(-\frac{1}{x\tan x+1}\right)dx\\ & =-\frac{x^2}{x\tan x+1}+2\int \frac{x\cos x}{x\sin x+\cos x}dx\\ & =-\frac{x^2}{x\tan x+1}+2{\log }_e|x\sin x+\cos x|+c\\ & \left(\because \frac{d}{dx}(x\sin x+\cos x)=x\cos x\right)\end{array}$