Tính tích phân I =( x^2 dx) / (x^2-3x+2)

* Hướng dẫn giải

Chọn C

Ta có 

$$\begin{gathered} \frac{x^2}{x^2-3x+2}=1+\frac{3}{2}\frac{2x-3}{x^2-3x+2}+\frac{5}{2}\frac{1}{x^2-3x+2} \\ =1+\frac{3}{2}\frac{2x-3}{x^2-3x+2}+\frac{5}{2}\left(\frac{1}{x-2}-\frac{1}{x-1}\right) \end{gathered}$$

Suy ra 

$$\begin{gathered} I=\left(x+\frac{3}{2}\ln \left|x^2-3x+2\right|+\frac{5}{2}\ln \left|\frac{x-2}{x-1}\right|\right){|}_3^4 \\ =1+\frac{3}{2}\ln 3+\frac{5}{2}\ln \frac{4}{3} \end{gathered}$$