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

* Hướng dẫn giải

Chọn D

Ta có: $x^3-3x+2={(x-1)}^2(x+2)$

$$\begin{gathered} 2x+3=a{(x-1)}^2+b(x+2)(x-1)+c(x+2) \\ \iff 2x+3=(a+b)x^2+(c-2a+b)x+a-2b+2c \end{gathered}$$

$$\begin{gathered} \iff \left\{\begin{array}{c}a+b=0\\ -2a+b+c=2\\ a-2b+2c=3\end{array}\right. \\ \iff a=-\frac{1}{9},b=\frac{1}{9},c=\frac{5}{3} \end{gathered}$$

$$\begin{gathered} J={\int }_2^3\left[-\frac{1}{9}\frac{1}{x+2}+\frac{1}{9}\frac{1}{x-1}+\frac{5}{3}\frac{1}{{\left(x-1\right)}^2}\right]dx \\ =\left(\frac{1}{9}\ln \left|\frac{x-1}{x+2}\right|-\frac{5}{3(x-1)}\right){|}_2^3=\frac{1}{9}\ln \frac{8}{5}+\frac{5}{6} \end{gathered}$$