Answer:
First, express the fraction in partial fractions
Write it out as an identity:
[tex]\dfrac{2x-10}{x(x+1)(x-1)} \equiv \dfrac{A}{x}+\dfrac{B}{(x+1)}+\dfrac{C}{(x-1)}[/tex]
Add the partial fractions:
[tex]\dfrac{2x-10}{x(x+1)(x-1)} \equiv \dfrac{A(x+1)(x-1)+Bx(x-1)+Cx(x+1)}{x(x+1)(x-1)}[/tex]
Cancel the denominators from both sides of the original identity, so the numerators are equal:
[tex]2x-10 \equiv A(x+1)(x-1)+Bx(x-1)+Cx(x+1)[/tex]
Now solve for A, B and C by substitution.
Substitute values of x which make one of the expressions equal zero (to eliminate all but one of A, B and C):
[tex]\begin{aligned}x=0 \implies 2(0)-10 & =A(0+1)(0-1)+B(0)(0-1)+C(0)(0+1)\\-10 & = -A\\\implies A & = 10\end{aligned}[/tex]
[tex]\begin{aligned}x=1 \implies 2(1)-10 & =A(1+1)(1-1)+B(1)(1-1)+C(1)(1+1)\\-8 & = 2C\\\implies C & = -4\end{aligned}[/tex]
[tex]\begin{aligned}x=-1 \implies 2(-1)-10 & =A(-1+1)(-1-1)+B(-1)(-1-1)+C(-1)(-1+1)\\-12 & = 2B\\\implies B & = -6\end{aligned}[/tex]
Replace the found values of A, B and C in the original identity:
[tex]\implies \dfrac{2x-10}{x(x+1)(x-1)} \equiv \dfrac{10}{x}-\dfrac{6}{(x+1)}-\dfrac{4}{(x-1)}[/tex]
Now integrate:
[tex]\begin{aligned}\displaystyle \int \dfrac{2x-10}{x(x+1)(x-1)}\:dx & =\int \dfrac{10}{x}-\dfrac{6}{(x+1)}-\dfrac{4}{(x-1)}\:\:dx\\\\& =10\int \dfrac{1}{x}\:dx\:\:-6 \int\dfrac{1}{(x+1)}\:dx\:\:-4 \int \dfrac{1}{(x-1)}\:dx\\\\& = 10 \ln |x|-6 \ln |x+1|-4 \ln |x-1|+C\end{aligned}[/tex]
