Question

$\int dy=\int \frac{1}{x\log x}dx+\int x{sin}^{2}xdx$

Answer 1

$\int dy=\int \frac{1}{xlogx}dx+\int x{\sin }^{2}xdx$

$y=I_1+I_2$

$I_1=\int \frac{1}{t}dt=lnt=ln\left(logx\right)$

$I_2=\int x{\sin }^{2}xdx$

$=\frac{1}{2}\int xdx-\frac{1}{2}\int x\cos 2xdx$

$=\frac{1}{2}\cdot \frac{x^2}{2}-\frac{1}{2}[x\frac{\sin 2x}{2}-\int 1-\frac{\sin 2x}{2}dx]$

$=\frac{1}{4}{x}^2-\frac{1}{4}x\sin 2x+\frac{1}{4}\left(\frac{(-\cos 2x}{2}\right)+c$

$=\frac{1}{4}{x}^2-\frac{1}{4}\times \sin 2x-\frac{1}{8}\cos 2x+c$

$y=log\left(logx\right)+\frac{1}{4}{x}^2-\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+c$.