Question

How do you integrate `int x^3/(x^2 + 4x + 3)` using partial fractions?

Answer 1

`int \ x^3/(x^2+4x+3) \ dx = 1/2x^2 -4x +27/2ln|x+3|-1/2ln|x+1| + c`

Explanation:

We seek:

`I = int \ x^3/(x^2+4x+3) \ dx`

Before we consider a partial fraction decomposition we note that the order of the numerator is one degree higher than the order of the denominator (the equivalent of a "top-heavy" fraction). So we must use algebraic long division in order to reduce the order.

The algebraic division is as follows

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ -4`
`x^2+4x+3 \ bar( ")" x^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ) \ -`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^3+4x^2+3x`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bar( \ \ 0-4x^2-3x \ \ \ \ \ \ \ \ \ \ \ ) \ \ -`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -4x^2-16x-12`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bar( \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ 13x+12 \ \ )`

And so we can write the integral as follows:

`I = int \ (x-4) + (13x+12)/(x^2+4x+3) \ dx`
`\ \ = int \ (x-4) \ dx + int \ (13x+12)/((x+3)(x+1)) \ dx`

We can readily handle the first integral, so now we must deal with the second integral by decomposing the integrand into partial fractions, which will take the form:

`(13x+12)/((x+3)(x+4)) -= A/(x+3) + B/(x+1)`
`" " = (A(x+1) + B(x+3))/((x+3)(x+1))`

Leading to the identity:

`13x+12 = A(x+1) + B(x+3)`

Where `A,B` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x = -3 => -27 = -2A => A = 27/2`
Put `x = -1 => -1 = 2B \ \ \ \ \ => B = -1/2`

So using partial fraction decomposition we have:

`I = int \ (x-4) \ dx + \ int \ (27/2)/(x+3) + (-1/2)/(x+1) \ dx`

And now all integrands are readily integratable, so:

`I = 1/2x^2 -4x +27/2ln|x+3|-1/2ln|x+1| + c`