How do you write the partial fraction decomposition of the rational expression  1/((x+6)(x^2+3))?

Mar 25, 2018

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{1}{39 \left(x + 6\right)} - \frac{x - 6}{39 \left({x}^{2} + 3\right)}$

Explanation:

Given:

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)}$

Note that ${x}^{2} + 3$ has no linear factors with real coefficients.

So, assuming we want a decomposition with real coefficients, it takes the form:

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{A}{x + 6} + \frac{B x + C}{{x}^{2} + 3}$

Multiplying both sides by $\left(x + 6\right) \left({x}^{2} + 3\right)$ this becomes:

$1 = A \left({x}^{2} + 3\right) + \left(B x + C\right) \left(x + 6\right)$

$\textcolor{w h i t e}{1} = \left(A + B\right) {x}^{2} + \left(6 B + C\right) x + \left(3 A + 6 C\right)$

Putting $x = - 6$ we get:

$1 = A \left({\left(\textcolor{b l u e}{- 6}\right)}^{2} + 3\right) = 39 A$

So:

$A = \frac{1}{39}$

Then from the coefficient of ${x}^{2}$, we have:

$A + B = 0$

So:

$B = - A = - \frac{1}{39}$

From the constant term we have:

$1 = 3 A + 6 C$

Hence:

$C = \frac{1}{6} \left(1 - 3 A\right) = \frac{1}{6} \left(1 - \frac{1}{13}\right) = \frac{2}{13} = \frac{6}{39}$

So:

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{1}{39 \left(x + 6\right)} - \frac{x - 6}{39 \left({x}^{2} + 3\right)}$