Integrating (2x^2 -3)/x(x^2 +4

• Mar 16th 2010, 06:36 PM
maxreality
Integrating (2x^2 -3)/x(x^2 +4
I'm stuck on this problem. I've been playing around, and I can't seem to figure out a way to start it. Is Heaviside the way to go about it or is there some easy simplification I'm not seeing?

$
\int \frac{2x^2-3}{x(x^2+4)}*dx
$
• Mar 16th 2010, 07:11 PM
Soroban
Hello, maxreality!

Quote:

$\int \frac{2x^2-3}{x(x^2+4)}\,dx$

I'd use Partial Fractions . . .

. . $\frac{2x^2-3}{x(x^2+4)} \;=\;\frac{A}{x} + \frac{Bx+C}{x^2+4}$

You should get: . $A \,=\,-\frac{3}{4},\;B \,=\,\frac{11}{4}$

• Mar 17th 2010, 07:14 AM
maxreality
Ah hah! I didn't realize Bx + C is for quadratics. Thanks!