\(\int \frac{x \mathrm{~d} x}{(x-1)^2(x+2)}=\)

Let \(I=\int \frac{x}{(x-1)^2(x+2)} \mathrm{d} x\)
Let \(\frac{x}{(x-1)^2(x+2)}=\frac{\mathrm{A}}{(x-1)}+\frac{\mathrm{B}}{(x-1)^2}+\frac{\mathrm{C}}{(x+2)}\)
\(\therefore \quad x=\mathrm{A}(x-1)(x+2)+\mathrm{B}(x+2)+\mathrm{C}(x-1)^2\)
for \(x=1\), we get \(\mathrm{B}=\frac{1}{3}\)
for \(x=-2\), we get \(\mathrm{C}=\frac{-2}{9}\)
Equating the coefficients of \(x^2\), we get
\(\mathrm{A}+\mathrm{C}=0\)
\(\therefore A=\frac{2}{9}\)
\(\therefore \mathrm{I} =\frac{2}{9} \int \frac{1}{(x-1)} \mathrm{d} x+\frac{1}{3} \int \frac{1}{(x-1)^2} \mathrm{~d} x-\frac{2}{9} \int \frac{1}{(x+2)} \mathrm{d} x \)
\( =\frac{2}{9} \log (x-1)-\frac{1}{3} \times \frac{1}{(x-1)}-\frac{2}{9} \log (x+2)+\mathrm{c}\)