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Evaluate $\int \frac{x^3}{(x-2)(x+3)} d x$.
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We start by writing $\frac{7 x-6}{(x-2)(x+3)}$ as the sum of two fractions. We want to end up with
$\frac{7 x-6}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3} .$
If we go ahead and add the fractions on the right hand side we get
$\frac{7 x-6}{(x-2)(x+3)}=\frac{(A+B) x+3 A-2 B}{(x-2)(x+3)} \text {. }$
So all we need to do is find $A$ and $B$ so that $7 x-6=(A+B) x+3 A-2 B$, which is to say, we need $7=A+B$ and $-6=3 A-2 B$. This is a problem you've seen before: solve a system of two equations in two unknowns. There are many ways to proceed; here's one: If $7=A+B$ then $B=7-A$ and so $-6=3 A-2 B=3 A-2(7-A)=3 A-14+2 A=5 A-14$.
This is easy to solve for $A: A=8 / 5$, and then $B=7-A=7-8 / 5=27 / 5$. Thus
$\int \frac{7 x-6}{(x-2)(x+3)} d x=\int \frac{8}{5} \frac{1}{x-2}+\frac{27}{5} \frac{1}{x+3} d x=\frac{8}{5} \ln |x-2|+\frac{27}{5} \ln |x+3|+C \text {. }$
The answer to the original problem is now
\begin{aligned} \int \frac{x^3}{(x-2)(x+3)} d x &=\int x-1 d x+\int \frac{7 x-6}{(x-2)(x+3)} d x \\ &=\frac{x^2}{2}-x+\frac{8}{5} \ln |x-2|+\frac{27}{5} \ln |x+3|+C \end{aligned}
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