Đặt $\ln x = u \Rightarrow \dfrac{1}{x}dx=du$ và $\dfrac{x}{(x^2+1)^2}dx=dv \Rightarrow \dfrac{1}{2}\dfrac{d(x^2+1)}{(x^2+1)^2}=dv$
$\Rightarrow -\dfrac{1}{2(x^2+1)}=v$
$I=-\dfrac{1}{2(x^2+1)}\ln x +\dfrac{1}{2} \int\dfrac{1}{x(x^2+1)}dx$
TÍnh $\int\dfrac{1}{x(x^2+1)}dx=\int \dfrac{x}{x^2(x^2+1)}dx =\dfrac{1}{2}\int \dfrac{d(x^2)}{x^2(x^2+1)}=\dfrac{1}{2}\int \dfrac{1}{t(t+1)}dt$
$=\dfrac{1}{2} \int \bigg (\dfrac{1}{t}-\dfrac{1}{t+1} \bigg )dt=\dfrac{1}{2}\ln \bigg |\dfrac{t}{t+1} \bigg | + C=\dfrac{1}{2}\ln \bigg |\dfrac{x^2}{x^2+1} \bigg | + C$
Tự ráp lại nhé