# How do you integrate int 1/(x^3 +4x) dx using partial fractions?

Feb 17, 2017

$\int \frac{\mathrm{dx}}{{x}^{3} + 4 x} = \frac{1}{8} \ln \left({x}^{2} / \left({x}^{2} + 4\right)\right) + C$

#### Explanation:

Factorize the denominator:

${x}^{3} + 4 x = x \left({x}^{2} + 4\right)$

So:

$\frac{1}{{x}^{3} + 4 x} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 4}$

$\frac{1}{{x}^{3} + 4 x} = \frac{A \left({x}^{2} + 4\right) + x \left(B x + C\right)}{x \left({x}^{2} + 4\right)}$

$\frac{1}{{x}^{3} + 4 x} = \frac{A {x}^{2} + 4 A + B {x}^{2} + C x}{x \left({x}^{2} + 4\right)}$

$\frac{1}{{x}^{3} + 4 x} = \frac{\left(A + B\right) {x}^{2} + C x + 4 A}{x \left({x}^{2} + 4\right)}$

and equating the coefficient with the same degree in $x$:

$\left\{\begin{matrix}A + B = 0 \\ C = 0 \\ 4 A = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{1}{4} \\ B = - \frac{1}{4} \\ C = 0\end{matrix}\right.$

We have then:

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

$\int \frac{\mathrm{dx}}{{x}^{3} + 4 x} = \frac{1}{4} \int \frac{\mathrm{dx}}{x} - \frac{1}{8} \int \frac{d \left({x}^{2} + 4\right)}{{x}^{2} + 4}$

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

that we can also write as:

$\int \frac{\mathrm{dx}}{{x}^{3} + 4 x} = \frac{1}{8} \ln \left({x}^{2} / \left({x}^{2} + 4\right)\right) + C$