# How can I integrate int 1/(x^4+x) dx using partial fractions?

## My problem is that I don't know what will be in the numerator over the ${x}^{3} + 1$ $A$ or $A x + B$? I've learnt that only these two options exist. But I also saw $A {x}^{2} + B$ so I'm kinda confused.

Dec 14, 2017

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

#### Explanation:

Now I decomposed integrand into basic fractions,

$\int \frac{\mathrm{dx}}{{x}^{4} + x}$

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

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

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

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

=$L n x - \frac{1}{3} L n \left({x}^{3} + 1\right) + C$

Dec 15, 2017

The answer is $= \ln \left(| x |\right) - \frac{1}{3} \ln \left(| x + 1 |\right) - \frac{1}{3} \ln \left(| {x}^{2} - x + 1 |\right) + C$

#### Explanation:

Reminder

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

The denominator is

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

Therefore, the decomposition into partial fractions is

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

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

The denominators are the same, compare the numerators

1=A(x+1)(x^2-x+1)+B(x)(x^2-x+1)+(Cx+D)(x)(x+1))

Compare the LHS and the RHS

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

Let $x = - 1$, $\implies$, $1 = B \left(- 1\right) \left(\right)$, $\implies$, $B = - \frac{1}{3}$

Coefficients of ${x}^{3}$

$0 = A + B + C$

$C = - A - B = - 1 + \frac{1}{3} = - \frac{2}{3}$

Coefficients of ${x}^{2}$

$0 = - B + C + D$

$D = B - C = - \frac{1}{3} + \frac{2}{3} = \frac{1}{3}$

Therefore,

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

So,

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

$= \ln \left(| x |\right) - \frac{1}{3} \ln \left(| x + 1 |\right) - \frac{1}{3} \ln \left(| {x}^{2} - x + 1 |\right) + C$