Question

$\int \frac{1}{x+xlogx}dx$.

Answer 1

${\int }^{}\frac{1}{x+x\log x}dx=\int \frac{1}{x(1+\log x)}dx$

Let $\log x=t\Rightarrow \frac{1}{x}dx=dt$

So, ${\int }^{}\frac{1}{x+x\log x}dx=\int \frac{dt}{1+t}=\ln |1+t|+C=\ln |1+\log x|+C$