# How do you integrate (3x+5)/(x^2+4x+13) using partial fractions?

Jun 14, 2017

You can't...

but the integral is $\frac{3}{2} \ln \left({x}^{2} + 4 x + 13\right) - \frac{1}{3} \arctan \left(\frac{x + 2}{3}\right) + C$

#### Explanation:

This integral cannot be solved with partial fractions, since the bottom polynomial cannot be factored. However, we can use the definition of the derivative of arctan(x) to solve this integral.

Let's first re-write the integral like this:

$\int \frac{3 x + 5}{{x}^{2} + 4 x + 13} \mathrm{dx} = \int \frac{3 x + 6 - 1}{{x}^{2} + 4 x + 13} \mathrm{dx}$

$= \int \frac{3 x + 6}{{x}^{2} + 4 x + 13} \mathrm{dx} - \int \frac{1}{{\left(x + 2\right)}^{2} + 9} \mathrm{dx}$

$= \frac{3}{2} \int \frac{2 x + 4}{{x}^{2} + 4 x + 13} \mathrm{dx} - \int \frac{1}{{\left(x + 2\right)}^{2} + 9} \mathrm{dx}$

For the first integral, we can use the substitution:

$u = {x}^{2} + 4 x + 13 \to \mathrm{du} = \left(2 x + 4\right) \mathrm{dx}$

$\therefore \int \frac{2 x + 4}{{x}^{2} + 4 x + 13} \mathrm{dx} = \int \frac{\mathrm{du}}{u} = \ln | u | = \ln | {x}^{2} + 4 x + 13 |$

For the second integral, we use the definition of the derivative of arctangent:

$\int \frac{\mathrm{dx}}{{x}^{2} + 1} = \arctan \left(x\right)$

And we also use this substitution:

$v = \frac{x + 2}{3} \to \mathrm{dv} = \frac{1}{3} \mathrm{dx} \to \mathrm{dx} = 3 \mathrm{dv}$

So now we get:

$\int \frac{1}{{\left(x + 2\right)}^{2} + 9} \mathrm{dx} = \int \frac{1}{{\left(3 v\right)}^{2} + 9} \cdot 3 \mathrm{dv} = \int \frac{1}{9 {v}^{2} + 9} \cdot 3 \mathrm{dv}$

$= \frac{1}{9} \cdot 3 \cdot \int \frac{\mathrm{dv}}{{v}^{2} + 1} = \frac{1}{3} \arctan \left(v\right) = \frac{1}{3} \arctan \left(\frac{x + 2}{3}\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Therefore, the complete integral becomes:

$\frac{3}{2} \int \frac{2 x + 4}{{x}^{2} + 4 x + 13} \mathrm{dx} - \int \frac{1}{{\left(x + 2\right)}^{2} + 9} \mathrm{dx}$

$= \frac{3}{2} \ln | {x}^{2} + 4 x + 13 | - \frac{1}{3} \arctan \left(\frac{x + 2}{3}\right) + C$

As a final adjustment, we can also get rid of the absolute value signs around ${x}^{2} + 4 x + 13$, since it is always positive for $x \in \mathbb{R}$.

$\frac{3}{2} \ln \left({x}^{2} + 4 x + 13\right) - \frac{1}{3} \arctan \left(\frac{x + 2}{3}\right) + C$