Question

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

Answer 1

`int(3x^2+2x)/((x+2)(x^2+4)) dx= ln(x+2)+ ln(x^2+4)−arctan(x/2)+C`

Explanation:

The partial fraction decomposition will be of the form:

`(3x^2+2x)/((x+2)(x^2+4)) -= A/(x+2) + (Bx+C)/(x^2+4)`
`:. (3x^2+2x)/((x+2)(x^2+4)) = (A(x^2+4) + (Bx+C)(x+2))/((x+2)(x^2+4))`
`:. 3x^2+2x = A(x^2+4) + (Bx+C)(x+2)`

We now need to find the three coefficients, `A`,`B` and `C`:

Put `x=-2=>12-4=A(4+4)+0`
`:. 8A=8 => A=1`

Equating coefficients of `x^2 => 3=A+B`
`:. B=3-1=2`

Equating coefficients of constants `=> 0=4A+2C`
`:. 2C=-4 => C=-2`

So, we have:
`(3x^2+2x)/((x+2)(x^2+4)) = 1/(x+2) + (2x-2)/(x^2+4)`

And, therefore,
`int(3x^2+2x)/((x+2)(x^2+4)) dx= int 1/(x+2) + (2x-2)/(x^2+4) dx`
`:. int(3x^2+2x)/((x+2)(x^2+4)) dx= int 1/(x+2) dx+ int(2x-2)/(x^2+4) dx`

`:. int(3x^2+2x)/((x+2)(x^2+4)) dx= ln(x+2)+ ln(x^2+4)−arctan(x/2)+C`

I have omitted the derivation of the second interval, as the question is about partial fractions and not integration by substation, you can use http://www.integral-calculator.com/ to get a full step by step.