Question

of partial fractions, compute
int \frac{10x^{2}-8x - 6}{x(x + 2)(x - 3)}dx

Explanation:

Step1: Decompose into partial - fractions

Let $\frac{10x^{2}-8x - 6}{x(x + 2)(x - 3)}=\frac{A}{x}+\frac{B}{x + 2}+\frac{C}{x - 3}$. Then $10x^{2}-8x - 6=A(x + 2)(x - 3)+Bx(x - 3)+Cx(x + 2)$.
If $x = 0$, then $-6=A(2)(-3)$, so $A = 1$.
If $x=-2$, then $10\times4+8\times2 - 6=B(-2)(-5)$, $40 + 16-6 = 10B$, $50 = 10B$, $B = 5$.
If $x = 3$, then $10\times9-8\times3 - 6=C(3)(5)$, $90-24 - 6=15C$, $60 = 15C$, $C = 4$. So $\frac{10x^{2}-8x - 6}{x(x + 2)(x - 3)}=\frac{1}{x}+\frac{5}{x + 2}+\frac{4}{x - 3}$.

Step2: Integrate term - by - term

$\int\frac{10x^{2}-8x - 6}{x(x + 2)(x - 3)}dx=\int(\frac{1}{x}+\frac{5}{x + 2}+\frac{4}{x - 3})dx=\int\frac{1}{x}dx+5\int\frac{1}{x + 2}dx+4\int\frac{1}{x - 3}dx$.
We know that $\int\frac{1}{u}du=\ln|u|+C$. So $\int\frac{1}{x}dx=\ln|x|$, $5\int\frac{1}{x + 2}dx=5\ln|x + 2|$, $4\int\frac{1}{x - 3}dx=4\ln|x - 3|$.

Answer:

$\ln|x|+5\ln|x + 2|+4\ln|x - 3|+C$