Question

integrate (x^2-3x-1)/(x^3+x^2-2x)

my answer is 1/2 ln x - 5/2 ln (x+2) + 3 ln (x-1)+C

is my answer correct?

Answer 1

\[\frac{x^2-3x-1}{x^3+x^2-2x}=\frac{x^2-3x-1}{x(x+2)(x-1)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-1}\]
\[x^2-3x-1=A(x+2)(x-1)+Bx(x-1)+Cx(x+2)\\
x^2-3x-1=(A+B+C)x^2+(A-B+2C)x-2A
\]
\[\begin{cases}A+B+C=1\\A-B+2C=-3\\-2A=-1\end{cases}~~\Rightarrow~~\begin{cases}A=\frac{1}{2}\\\\B=\frac{3}{2}\\\\C=-1\end{cases}\]
So you have
\[\frac{1}{2}\int\frac{dx}{x}+\frac{3}{2}\int\frac{dx}{x+2}-\int\frac{dx}{x-1}\]
The first term in your solution is correct, but the other two are not.