Question

How do you integrate `x^2/ [x^2+x+4]` using partial fractions?

Answer 1

`int frac{x^2}{x^2+x+4} "d"x`

`= x - 1/2 ln(x^2+x+4) + 7/sqrt15 tan^{-1}(frac{2x + 1}{sqrt15}) + c`

where `c` is the constant of integration.

Explanation:

First, write out the partial fractions

`frac{x^2}{x^2+x+4} = frac{x^2+(x+4) - (x+4)}{x^2+x+4}`

`= 1 - frac{x+4}{x^2+x+4}`

Fractions of the form `(f'(x))/f(x)` integrate to become `ln(f(x))`. Therefore, suppose `f(x) = x^2 + x +4`, then

`f'(x) = 2x + 1`

Rewriting the partial fraction,

`frac{x^2}{x^2+x+4} = 1 - 1/2 frac{2(x+4)}{x^2+x+4}`

`= 1 - 1/2 frac{2x+8}{x^2+x+4}`

`= 1 - 1/2 frac{2x+1}{x^2+x+4} + frac{7}{2} frac{1}{x^2+x+4}`

The last bit requires completing the square, as fractions of the form `1/((x-b)^2 +a^2)` integrate to form `1/a tan^{-1}((x-b)/a)`.

`x^2 + x + 4 = (x + 1/2)^2 + 15/4`

Rewriting the partial fraction,

`frac{x^2}{x^2+x+4} = 1 - 1/2 frac{2x+1}{x^2+x+4} + frac{14}{(2x + 1)^2 + 15}`

Now, performing the integration

`int frac{x^2}{x^2+x+4} "d"x = int "d"x - 1/2 int frac{2x+1}{x^2+x+4} "d"x`

`+ 7/2 int frac{1}{(x + 1/2)^2 + 15/4} "d"x`

`= x - 1/2 ln(x^2+x+4) + 7/sqrt15 tan^{-1}(frac{2x + 1}{sqrt15}) + c`

where `c` is the constant of integration.