# Integration of Rational Functions by Partial Fractions

1. Mar 20, 2012

### Jgoshorn1

1. The problem statement, all variables and given/known data

∫(x3+4)/(x2+4)dx

2. Relevant equations

n/a

3. The attempt at a solution

I know I have to do long division before I can break this one up into partial fractions. So I x3+4 by x2+4 and got x with a remainder of -4x+4 to be written as x+(4-4x/x2+4).

Then I rewrote the initial integral as ∫x + ∫(4-4x)/(x2+4) but then I get stuck from there.
Should the partial fraction be ∫x + ∫(Ax+B)/(x2+4)?
And if so, how would I go about solving that?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 20, 2012

### LCKurtz

$\frac x {x^2+4}$ can be done with a u-substitution and $\frac 2 {x^2+4}$is an Arctan form.

3. Mar 20, 2012

### Jgoshorn1

How/where are you getting x/x2+4 and 2/x2+4 from?

4. Mar 20, 2012

### SithsNGiggles

Long division for
$\frac{x^3+4}{x^2+4}$

yields
$x - 4(\frac{x-1}{x^2+4})$.

As you mentioned, you now have
$\int x dx - 4\int \frac{x-1}{x^2+4} dx$.

For the second integral, split up the integrand:
$\int x dx - 4(\int \frac{x}{x^2+4} dx - \int \frac{1}{x^2+4} dx)$.

(This is what LCKurtz did, at least.)

Last edited: Mar 20, 2012
5. Mar 21, 2012

### Jgoshorn1

So do you think it was just a typo that he wrote 2/x2+4 instead of 1/x2+4 because that 2 really confused me

6. Mar 21, 2012

### LCKurtz

Yes, the 2 was a typo. At any rate a constant factor wouldn't affect the method of integration.