# How do you integrate (6x)/(x^3-8) using partial fractions?

##### 1 Answer
Jan 8, 2017

The answer is =ln(∣x-2∣)-1/2ln(∣x^2+2x+4∣)+4/sqrt3arctan((x+1)/sqrt3)+C

#### Explanation:

The denominator is

${x}^{3} - 8 = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

Therefore,

$\frac{6 x}{{x}^{3} - 8} = \frac{6 x}{\left(x - 2\right) \left({x}^{2} + 2 x + 4\right)}$

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

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

So,

$6 x = A \left({x}^{2} + 2 x + 4\right) + \left(B x + C\right) \left(x - 2\right)$

Let $x = 2$, $\implies$, $12 = 12 A$, $\implies$, $A = 1$

Let $x = 0$, $\implies$, $0 = 4 A - 2 C$, $\implies$, $C = 2$

Coefficients of ${x}^{2}$, $\implies$, $0 = A + B$, $\implies$, $B = - 1$

Therefore,

$\frac{6 x}{{x}^{3} - 8} = \frac{1}{x - 2} + \frac{- x + 2}{{x}^{2} + 2 x + 4}$

So,

$\int \frac{6 x \mathrm{dx}}{{x}^{3} - 8} = \int \frac{\mathrm{dx}}{x - 2} - \int \frac{\left(x - 2\right) \mathrm{dx}}{{x}^{2} + 2 x + 4}$

We integrate separately

intdx/(x-2)=ln(∣x-2∣)

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

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

Let $u = {x}^{2} + 2 x + 4$, $\implies$, $\mathrm{du} = \left(2 x + 2\right) \mathrm{dx}$

Therefore,

$\int \frac{\left(x + 2\right) \mathrm{dx}}{{x}^{2} + 2 x + 4} = \frac{1}{2} \int \frac{\mathrm{du}}{u} = \frac{1}{2} \ln u$

=1/2ln(∣x^2+2x+4∣)

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

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

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

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

Let $\tan \theta = \frac{x + 1}{\sqrt{3}}$$\implies$ ${\sec}^{2} \theta d \theta = \frac{\mathrm{dx}}{\sqrt{3}}$

Therefore,

$\frac{4}{3} \int \frac{\mathrm{dx}}{\left({\left(\frac{x + 1}{\sqrt{3}}\right)}^{2} + 1\right)} = \frac{4}{3} \int \frac{\sqrt{3} {\sec}^{2} \theta d \theta}{1 + {\tan}^{2} \theta}$

But $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$

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

$= \frac{4}{\sqrt{3}} \arctan \left(\frac{x + 1}{\sqrt{3}}\right)$