Tích phân từng phần nha
Đặt $\ln (x+1) = u \Rightarrow \dfrac{1}{x+1} dx = du$
$\dfrac{dx}{x^2} = dv \Rightarrow -\dfrac{1}{x} = v$
$I = -\dfrac{1}{x} .\ln (1+x) + \int \dfrac{1}{x(x+1)} dx$
$= -\dfrac{1}{x} .\ln (1+x) + \int \bigg (\dfrac{1}{x} - \dfrac{1}{x+1} \bigg )dx$
$= -\dfrac{1}{x} .\ln (1+x) + \int \dfrac{dx}{x} - \int \dfrac{dx}{x+1}$
$= -\dfrac{1}{x} .\ln (1+x) + \ln |x| - \ln |x+1| +C$
$= -\dfrac{1}{x} .\ln (1+x) + \ln |\dfrac{x}{x+1}| +C$