Question

How do you use partial fraction decomposition to decompose the fraction to integrate `x/(x^2+4x+13)`?

Answer 1

This function should not be integrated with partial fractions, but with a substitution (after completing the square in the denominator). The final answer is `int x/(x^2+4x+13)\ dx`

`=1/2 ln(x^2+4x+13)-2/3 arctan((x+2)/3)+C`

Explanation:

The denominator cannot be factored without resorting to complex numbers, so we complete the square in the denominator instead.

Completing the square for the denominator `x^2+4x+13` can be done by taking the coefficient 4 of `x`, dividing it by 2 and squaring that to get 4, and then writing the denominator as `x^2+4x+4+13-4=(x^2+4x+4)+9=(x+2)^2+9`.

The integral can therefore be written as `\int x/(x^2+4x+13)\ dx=\int x/((x+2)^2+9)\ dx`.

Now let `u=x+2` so that `x=u-2` and `dx=du` to write

`\int x/((x+2)^2+9)\ dx=\int u/(u^2+9)\ du-\int 2/(u^2+9)\ du`

The first of these integrals can be done by inspection to see that `\int u/(u^2+9)\ du=1/2 ln|u^2+9|+C=1/2 ln(x^2+4x+13)+C` (you could also do another substitution `w=u^2+9`, `dw=2u\ du`, and `u\ du = 1/2 dw` to help you do this). Note also that `x^2+4x+13=(x+2)^2+9>0` for all `x\in RR`.

The second integral can be done by a bit of tricky algebra, motivated by the fact that `\int 1/(1+x^2)\ dx=arctan(x)+C`. Here's the algebra:

`\int 2/(u^2+9)\ du=2/9 \int 1/(1+(u/3)^2)\ du=2/3 arctan(u/3)+C`

`=2/3 arctan((x+2)/3)+C` (you could also do another substitution `w=u/3`, `du=3dw` to help you do this). Note at the end that we have replaced `u` by `x+2`.

Hence, after combining the integration constants into just one, the final answer is

`int x/(x^2+4x+13)\ dx`

`=1/2 ln(x^2+4x+13)-2/3 arctan((x+2)/3)+C`