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Evaluate \(\int \frac{x^3}{(x-2)(x+3)} d x\).
in Mathematics by Platinum (141,884 points) | 16 views

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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
\[
\begin{aligned}
\int \frac{x^3}{(x-2)(x+3)} d x &=\int x-1 d x+\int \frac{7 x-6}{(x-2)(x+3)} d x \\
&=\frac{x^2}{2}-x+\frac{8}{5} \ln |x-2|+\frac{27}{5} \ln |x+3|+C
\end{aligned}
\]
by Platinum (141,884 points)

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