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

**maxreality** · March 16th 2010, 05:36 PM

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?

Thanks in advance,

$<br /> \int \frac{2x^2-3}{x(x^2+4)}*dx<br />$

**Soroban** · March 16th 2010, 06:11 PM

Hello, maxreality!

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

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

**maxreality** · March 17th 2010, 06:14 AM

Ah hah! I didn't realize Bx + C is for quadratics. Thanks!

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