Integrate the following function:

$\frac{1}{x(x^4-1)}$

Question

Integrate the following function:

$\frac{1}{x(x^4-1)}$

Accepted Answer

> [!TLDR]
> Factor $x^4-1=(x-1)(x+1)(x^2+1)$, decompose, and integrate: $-\log|x|+\dfrac14\log|x-1|+\dfrac14\log|x+1|+\dfrac14\log(x^2+1)+C$.

## Factor and set up

$$x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1),$$
so the denominator is $x(x-1)(x+1)(x^2+1)$: three distinct linear factors and one irreducible quadratic. The irreducible quadratic gets a linear numerator:
$$\frac{1}{x(x-1)(x+1)(x^2+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{Dx+E}{x^2+1}.$$

## Solve for the constants

Clearing denominators,
$$1=A(x^4-1)+B\,x(x+1)(x^2+1)+C\,x(x-1)(x^2+1)+(Dx+E)\,x(x^2-1).$$
Substitute the real roots to get the linear constants quickly:

- $x=0$: $\;1=A(-1)\Rightarrow A=-1$
- $x=1$: $\;1=B\,(1)(2)(2)\Rightarrow B=	frac14$
- $x=-1$: $\;1=C\,(-1)(-2)(2)\Rightarrow C=	frac14$

For $D,E$, compare the highest and $x^3$ coefficients. The $x^4$ terms give $A+B+C+D=0\Rightarrow -1+	frac14+	frac14+D=0\Rightarrow D=	frac12$. The $x^3$ terms give $B-C+E=0\Rightarrow E=0$. So
$$A=-1,\quad B=	frac14,\quad C=	frac14,\quad D=	frac12,\quad E=0.$$

## The decomposed form

$$\frac{1}{x(x^4-1)}=-\frac1x+\frac{1/4}{x-1}+\frac{1/4}{x+1}+\frac{(1/2)x}{x^2+1}.$$

## Integrate term by term …

Q.Integrate the following function: $\frac{1}{x(x^4-1)}$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Factor and set up

Solve for the constants

The decomposed form

Integrate term by term …

Method: Full factorisation with a linear numerator over the irreducible quadratic

Steps

Common Mistakes

Q.Integrate the following function: 1x(x4−1)\frac{1}{x(x^4-1)}x(x4−1)1​

Factor and set up

Solve for the constants

The decomposed form

Integrate term by term …

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Q.Integrate the following function: $\frac{1}{x(x^4-1)}$

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