# Math Help - integration by partial fractions

1. ## integration by partial fractions

i need to take the integral of: x/(x^2 + 4 * x + 13) using partial fractions. my teacher told me to start off by turning this into the integral of: x/(x + 2)^2 +9, but i have no idea what to do when i get it in that form, anyone care to shed some light on this for me?

2. $\int {\frac{x}
{{x^2 + 4x + 13}}\,dx} = \frac{1}
{2}\int {\frac{{2x}}
{{x^2 + 4x + 13}}\,dx}.$
So

$\int {\frac{x}
{{x^2 + 4x + 13}}\,dx} = \frac{1}
{2}\left( {\int {\frac{{2x + 4}}
{{x^2 + 4x + 13}}\,dx} - \int {\frac{4}
{{(x + 2)^2 + 9}}\,dx} } \right).$

The first integral is immediate, now take a look to the second one, it's an arctangent, can you do that?

3. ok, so i i have been trying to work through this and now i am at:

integral of x/(x+2)^2 +9 , i see that if i can set u = to x+2 i can use the arcsin thing, however i have that stupid x on top that won't allow me to use that substitution, can anyone help me out?

4. Krizalid's method is probably the best for this integral, but if you want to practise integrating after partial fractions (which is very useful) then the following may be useful:
u = x+2 => x = u-2
$
\frac {u-2}{u^2} = \frac {u}{u^2} - \frac {2}{u^2}$

Good job finding the right substitution to use.

$\int \frac{x}{(x + 2)^2 + 9} dx = \int \frac{x}{9((\frac{x + 2}{3})^2 + 1)} dx = \tfrac{1}{9} \int \frac{x}{(\frac{x + 2}{3})^2 + 1} dx$

$\tan \theta = \frac{x+2}{3}$
$\sec^2 \theta d \theta = \tfrac{1}{3} dx$

$= \tfrac{1}{9} \int \frac{3\overbrace{(3\tan \theta - 2)}^{x} \sec^2 \theta}{(\tan \theta)^2 + 1} d \theta = \tfrac{1}{3} \int \frac{(3 \tan \theta - 2) \sec^2 \theta}{\sec^2 \theta} d \theta = \tfrac{1}{3} \int (3 \tan \theta - 2) d \theta$

$= \tfrac{1}{3} \Big[ 3 \int \tan \theta d \theta - 2 \theta \Big] = \tfrac{1}{3} \Big[ 3 \int \tan \theta d \theta - 2 \theta \Big] = - \tfrac{1}{3} \Big[3ln|\cos (\tfrac{x + 2}{3})| + 2 \arctan (\tfrac{x + 2}{3}) \Big]$

$= \boxed{- \tfrac{1}{3} \Big[3 ln|\tfrac{3}{\sqrt{x^2 + 4x + 13}}| + 2 \arctan (\tfrac{x + 2}{3}) \Big]}$

I probably made a mistake somewhere but that seems like the general idea.