# How do you integrate (x+4)/[(x+1)^2 + 4] using partial fractions?

Apr 18, 2018

$\int \frac{x + 4}{{\left(x + 1\right)}^{2} + 4} \mathrm{dx} = \ln \sqrt{{\left(x + 1\right)}^{2} + 4} + \frac{3}{2} \arctan \left(\frac{x + 1}{2}\right) + C$

#### Explanation:

There is no way to really perform partial fraction decomposition here; however, we may integrate using the following substitution:

$u = x + 1$
$x = u - 1$
$\mathrm{du} = \mathrm{dx}$

$\int \frac{u - 1 + 4}{{u}^{2} + 4} \mathrm{dx} = \int \frac{u + 3}{{u}^{2} + 4} \mathrm{du}$

We may split this up as follows:

$\int \frac{u}{{u}^{2} + 4} \mathrm{du} + 3 \int \frac{\mathrm{du}}{{u}^{2} + 4}$

$\int \frac{u}{{u}^{2} + 4} \mathrm{du} = \frac{1}{2} \ln \left({u}^{2} + 4\right)$ -- this can be solved with a mental substitution, as the differential of ${u}^{2} + 4$ shows up in the numerator with a bit of simplification. We do not attach absolute value bars as ${u}^{2} + 4$ is always positive.

$3 \int \frac{\mathrm{du}}{{u}^{2} + 4} = \frac{3}{2} \arctan \left(\frac{u}{2}\right)$.

This comes from the common integral, $\int \frac{\mathrm{dx}}{{x}^{2} + {a}^{2}} = \frac{1}{a} \arctan \left(\frac{x}{a}\right)$. For our case, $a = \sqrt{4} = 2$.

Thus, rewriting in terms of $x$ and simplifying the logarithm, we see

$\int \frac{x + 4}{{\left(x + 1\right)}^{2} + 4} \mathrm{dx} = \ln \sqrt{{\left(x + 1\right)}^{2} + 4} + \frac{3}{2} \arctan \left(\frac{x + 1}{2}\right) + C$