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

Jun 25, 2016

With Real coefficients:

$\frac{x + 4}{{\left(x + 1\right)}^{2} + 4} = \frac{x + 4}{{x}^{2} + 2 x + 5}$

With Complex coefficients:

$\frac{x + 4}{{\left(x + 1\right)}^{2} + 4} = \frac{2 - 3 i}{4 \left(x + 1 - 2 i\right)} + \frac{2 + 3 i}{4 \left(x + 1 + 2 i\right)}$

#### Explanation:

Note that if $x$ is Real then ${\left(x + 1\right)}^{2} \ge 0$ so ${\left(x + 1\right)}^{2} + 4 \ge 4 > 0$.

Hence: ${\left(x + 1\right)}^{2} + 4$ has no linear factors.

So if we are restricted to Real coefficients then this does not break down into simpler partial fractions.

We can expand and render the denominator in standard form:

$\frac{x + 4}{{\left(x + 1\right)}^{2} + 4} = \frac{x + 4}{{x}^{2} + 2 x + 5}$

On the other hand, if we want to use Complex coefficients then the denominator factors as a difference of squares:

${\left(x + 1\right)}^{2} + 4 = {\left(x + 1\right)}^{2} - {\left(2 i\right)}^{2} = \left(x + 1 - 2 i\right) \left(x + 1 + 2 i\right)$

So for some $A , B \in \mathbb{C}$ we have:

$\frac{x + 4}{{\left(x + 1\right)}^{2} + 4} = \frac{A}{x + 1 - 2 i} + \frac{B}{x + 1 + 2 i}$

$= \frac{A \left(x + 1 + 2 i\right) + B \left(x + 1 - 2 i\right)}{{\left(x + 1\right)}^{2} + 4}$

$= \frac{\left(A + B\right) x + \left(A \left(1 + 2 i\right) + B \left(1 - 2 i\right)\right)}{{\left(x + 1\right)}^{2} + 4}$

Equating coefficients:

$\left\{\begin{matrix}A + B = 1 \\ A \left(1 + 2 i\right) + B \left(1 - 2 i\right) = 4\end{matrix}\right.$

Subtracting the first equation from the second we get:

$2 i \left(A - B\right) = 3$

So:

$A - B = \frac{3}{2 i} = - \frac{3}{2} i$

Hence:

$A = \frac{1}{2} - \frac{3}{4} i$

$B = \frac{1}{2} + \frac{3}{4} i$

So:

$\frac{x + 4}{{\left(x + 1\right)}^{2} + 4} = \frac{2 - 3 i}{4 \left(x + 1 - 2 i\right)} + \frac{2 + 3 i}{4 \left(x + 1 + 2 i\right)}$