Ta có \(\int_{1}^{2}\frac{x^2+3x+3}{x(x^2+4x+3)}dx=\int_{1}^{2}\frac{dx}{x}-\int_{1}^{2}\frac{dx}{x^2+4x+3}\) \(=\int_{1}^{2}\frac{dx}{x}-\frac{1}{2}(\frac{1}{x+1}-\frac{1}{x+3})dx\) Suy ra \(I=ln\left | x \right |\bigg|_{1}^{2}-\frac{1}{2}ln\left | \frac{x+1}{x+2} \right | \bigg |_{1}^{2}= ln2 - \frac{1}{2}ln\frac{6}{5}\)
