# How do you integrate (4x^3+2x^2+1)/(4x^3-x) using partial fractions?

Jan 20, 2018

The answer is $= x - \ln \left(| x |\right) + \frac{1}{2} \ln \left(| 2 x + 1 |\right) + \ln \left(| 2 x - 1 |\right) + C$

#### Explanation:

As the degree of the numerator is equal to the degree of the denominator, perform a polynomial long division

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

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

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

Perform the partial fraction decomposition

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

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

The denominators are the same, compare the numerators

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

Let $x = 0$, $\implies$, $1 = - A$, $\implies$, $A = - 1$

Let $x = - \frac{1}{2}$, $\implies$, $1 = B$

Let $x = \frac{1}{2}$, $\implies$, $C = 2$

Therefore,

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

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

$= x - \ln \left(| x |\right) + \frac{1}{2} \ln \left(| 2 x + 1 |\right) + \ln \left(| 2 x - 1 |\right) + C$