# How do you integrate int (x^3 + 2) / (x^4 -16) using partial fractions?

Jun 6, 2016

$\frac{3}{16} \ln \left\mid x + 2 \right\mid + \frac{5}{16} \ln \left\mid x - 2 \right\mid + \frac{1}{4} \ln \left({x}^{2} + 4\right) - \frac{1}{8} \arctan \left(\frac{x}{2}\right) + K$

#### Explanation:

Firstly, you need the $\mathrm{dx}$ at the end of the integral for formal presentation i.e. $\int \frac{{x}^{3} + 2}{{x}^{4} - 16} \mathrm{dx}$.

Secondly, you need to decompose $\frac{{x}^{3} + 2}{{x}^{4} - 16}$ into partial fractions. This is made simpler by noting that

${x}^{4} - 16 = \left({x}^{2} - 4\right) \left({x}^{2} + 4\right) = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 4\right)$

That is,

$\frac{{x}^{3} + 2}{{x}^{4} - 16} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{{x}^{2} + 4}$

for real constants $A$, $B$, $C$ and $D$.

Using the cover-up rule and comparison of coefficients by algebraic means, you should be able to find solve for the unknowns, that is, $A = \frac{3}{16}$, $B = \frac{5}{16}$, $C = \frac{1}{2}$ and D=–1/4.

Thus, $\frac{{x}^{3} + 2}{{x}^{4} - 16} = \frac{\frac{3}{16}}{x + 2} + \frac{\frac{5}{16}}{x - 2} + \frac{\frac{1}{2} x - \frac{1}{4}}{{x}^{2} + 4}$, and

$\int \frac{{x}^{3} + 2}{{x}^{4} - 16} \mathrm{dx} = \int \left(\frac{\frac{3}{16}}{x + 2} + \frac{\frac{5}{16}}{x - 2} + \frac{\frac{1}{2} x - \frac{1}{4}}{{x}^{2} + 4}\right) \mathrm{dx}$.

The first two integrals are simple, i.e.

$\int \frac{\frac{3}{16}}{x + 2} \mathrm{dx} = \frac{3}{16} \int \frac{1}{x + 2} \mathrm{dx} = \frac{3}{16} \ln \left\mid x + 2 \right\mid + {K}_{1}$

and likewise,

$\int \frac{\frac{5}{16}}{x + 2} \mathrm{dx} = \frac{5}{16} \ln \left\mid x - 2 \right\mid + {K}_{2}$

The third part is a little tricky. By appropriate decompositions,

int (1/2x-1/4)/(x^2+4) dx = 1/4 int (2x)/(x^2+4) dx - 1/4 int 1/(x^2+4) dx = 1/4 ln(x^2+4) - 1/4(1/2)arctan (x/2) + K_3 = 1/4 ln(x^2+4) - 1/8 arctan (x/2) + K_3

Thus, putting everything together,

$\int \frac{{x}^{3} + 2}{{x}^{4} - 16} \mathrm{dx} = \frac{3}{16} \ln \left\mid x + 2 \right\mid + \frac{5}{16} \ln \left\mid x - 2 \right\mid + \frac{1}{4} \ln \left({x}^{2} + 4\right) - \frac{1}{8} \arctan \left(\frac{x}{2}\right) + K$

where $K = {K}_{1} + {K}_{2} + {K}_{3}$.