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

Integrate the following function: (1)/(x(x⁴-1))

Exercise 7.5 · Q20

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

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Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

Adding $\frac{2}{x-1}+\frac{3}{x+2}$ over a common denominator gives $\frac{5x+1}{x^2+x-2}$ . Partial fraction decomposition reverses this — it breaks one complicated rational function back into a sum of simple pieces. Those pieces are far easier to integrate: $\int \frac{2}{x-1}\,dx = 2\log|x-1|$ is immediate, while the combined fraction is not.

When it applies

You need a proper rational function, $\deg P < \deg Q$ . If the numerator's degree is equal or higher, first do polynomial long division. The denominator $Q(x)$ must be factored into linear and/or irreducible quadratic factors.

The standard forms

For $\frac{P(x)}{Q(x)}$ with $Q$ fully factored, each factor contributes a term:

Distinct linear $(ax+b)$ $\to$ $\dfrac{A}{ax+b}$ .
Repeated linear $(ax+b)^n$ $\to$ $\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+\cdots+\dfrac{A_n}{(ax+b)^n}$ .
Irreducible quadratic $(ax^2+bx+c)$ $\to$ $\dfrac{Ax+B}{ax^2+bx+c}$ — a linear numerator, not just a constant.

The unknown constants are found from the resulting equations; the decomposition is unique, which is what lets us solve for them systematically.

Worked example

Decompose $\frac{3x+5}{x^2-3x+2}$ . Factor the denominator: $(x-1)(x-2)$ . Set

\frac{3x+5}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2}.

Clear denominators: $3x+5 = A(x-2)+B(x-1)$ . Substituting the roots, $x=1$ gives $8=-A$ so $A=-8$ , and $x=2$ gives $11=B$ . Hence

\frac{3x+5}{x^2-3x+2} = \frac{-8}{x-1}+\frac{11}{x-2}.

Tip

With distinct linear factors, substitute each factor's root to knock out all but one term — much faster than equating coefficients. …

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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:

$x=0$ : $\;1=A(-1)\Rightarrow A=-1$
$x=1$ : $\;1=B\,(1)(2)(2)\Rightarrow B=\tfrac14$
$x=-1$ : $\;1=C\,(-1)(-2)(2)\Rightarrow C=\tfrac14$

For $D,E$ , compare the highest and $x^3$ coefficients. The $x^4$ terms give $A+B+C+D=0\Rightarrow -1+\tfrac14+\tfrac14+D=0\Rightarrow D=\tfrac12$ . The $x^3$ terms give $B-C+E=0\Rightarrow E=0$ . So

A=-1,\quad B=\tfrac14,\quad C=\tfrac14,\quad D=\tfrac12,\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 …

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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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