Integrate the rational function: $\frac{5x}{(x+1)(x^2-4)}$

(N/A) Given the integral: $I = \int \frac{5x}{(x+1)(x^2-4)} dx$.
First,factor the denominator: $(x+1)(x^2-4) = (x+1)(x+2)(x-2)$.
Using partial fractions,let $\frac{5x}{(x+1)(x+2)(x-2)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x-2}$.
Then,$5x = A(x+2)(x-2) + B(x+1)(x-2) + C(x+1)(x+2)$.
Setting $x = -1$: $5(-1) = A(1)(-3) \Rightarrow -5 = -3A \Rightarrow A = \frac{5}{3}$.
Setting $x = -2$: $5(-2) = B(-1)(-4) \Rightarrow -10 = 4B \Rightarrow B = -\frac{5}{2}$.
Setting $x = 2$: $5(2) = C(3)(4) \Rightarrow 10 = 12C \Rightarrow C = \frac{5}{6}$.
Thus,$I = \int \left( \frac{5/3}{x+1} - \frac{5/2}{x+2} + \frac{5/6}{x-2} \right) dx$.
Integrating term by term,we get $I = \frac{5}{3} \ln|x+1| - \frac{5}{2} \ln|x+2| + \frac{5}{6} \ln|x-2| + K$,where $K$ is the constant of integration.