Integration of Rational Functions by Partial Fractions

[Jgoshorn1]

Homework Statement

∫(x³+4)/(x²+4)dx

Homework Equations

n/a

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?

 

[LCKurtz]

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

 

[Jgoshorn1]

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

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

 

[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.)

 

[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

 

[LCKurtz]

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

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