Đặt \(\left\{{}\begin{matrix}u=ln\left(ax+b\right)\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{a}{ax+b}dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I=x.ln\left(ax+b\right)-\int\dfrac{ax}{ax+b}dx=x.ln\left(ax+b\right)-\int\left(1-\dfrac{b}{ax+b}\right)dx\)
\(=x.ln\left(ax+b\right)-x+\dfrac{b}{a}ln\left(ax+b\right)+C\)