Question

integrate the function [1/(x + xlogx)].dx

Answer 1

$\bf{I=\int{\frac{1}{x+xlogx}}\,dx}\\\\=\bf{\int{\frac{1}{x(1+logx)}}\,dx}\\\\=\bf{\int{\frac{1}{x}\frac{1}{1+logx}}\,dx}$

Let (1 + logx) = f(x) ------(1)

differentiate both sides,

1/x = f'(x) -------(2)

put equations (1) and (2) in I,

I = $\bf{\int{\frac{f'(x)}{f(x)}}\,dx}$

= $\bf{log_e|f(x)|}+C$

put f(x) = (1 + logx) ,

I = $\bf{log_e|1 + logx|}+C$