Partial Fractions: (x³ + 1)/(x² + 4)²

(x³ + 1)/(x² + 4)² decomposes into x/(x^2 + 4) + (-4x + 1)/(x^2 + 4)^2. A worked repeated quadratic example with the factorisation, the coefficient system, and an editable calculator.

(x³ + 1)/(x² + 4)² = x/(x^2 + 4) + (-4x + 1)/(x^2 + 4)^2. A repeated irreducible quadratic needs a linear numerator over each power.

Numerator N(x)

e.g. 3x+11

Denominator D(x)

Expanded or factored, e.g. (x-2)(x+3) or x^2+x-6

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The decomposition

(x³ + 1)/(x² + 4)² = x/(x^2 + 4) + (-4x + 1)/(x^2 + 4)^2

Denominator factors: (x^2 + 4)^2

This is a repeated quadratic problem. A repeated irreducible quadratic needs a linear numerator over each power.

All worked examples

1/(x² − 1) (3x + 11)/(x² − x − 6) 1/(x(x + 1)(x + 2)) (5x − 4)/(2x² + x − 1) 1/(x + 1)² x²/(x − 1)³ 8/(x³ + 2x² + x) (x² + 1)/(x³ + x) 1/(x³ − 1) (2x + 3)/((x + 1)(x² + 4)) 1/(x⁴ − 1) x³/(x² − 1) (x² + 2x)/((x + 1)(x − 2))

Frequently asked questions

How does (x³ + 1)/(x² + 4)² decompose? ▾

(x³ + 1)/(x² + 4)² = x/(x^2 + 4) + (-4x + 1)/(x^2 + 4)^2. A repeated irreducible quadratic needs a linear numerator over each power.

How was the denominator factored? ▾

It factors as (x^2 + 4)^2, which sets the shape of the decomposition.

How do I check the answer? ▾

Add the resulting fractions back over a common denominator — you should recover the original expression. Substituting a convenient value of x into both sides is a quicker spot check.

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