\(I=\int\limits^4_0x.ln\left(x^2+9\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=ln\left(x^2+9\right)\\dv=x.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{2x}{x^2+9}dx\\v=\dfrac{x^2}{2}\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{x^2}{2}ln\left(x^2+9\right)|^4_0-\int\limits^4_0\dfrac{x^3}{x^2+9}dx=16ln5-I_1\)
\(I_1=\int\limits^4_0\dfrac{x^3}{x^2+9}dx=\int\limits^4_0\left(x-\dfrac{9x}{x^2+9}\right)dx=\int\limits^4_0x.dx-\dfrac{9}{2}\int\limits^4_0\dfrac{d\left(x^2+9\right)}{x^2+9}\)
\(=\dfrac{x^2}{2}|^4_0-\dfrac{9}{2}ln\left(x^2+9\right)|^4_0=8-9ln5+9ln3\)
\(\Rightarrow I=16ln5-8+9ln5-9ln3=25ln5-9ln3-8\)
