Partial Fractions: 1/(x + 1)²

1/(x + 1)² decomposes into 1/(x + 1)^2. A worked repeated linear factor example with the factorisation, the coefficient system, and an editable calculator.

1/(x + 1)² = 1/(x + 1)^2. A repeated factor needs one term per power: A/(x + 1) plus B/(x + 1)².

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

1/(x + 1)² = 1/(x + 1)^2

Denominator factors: (x + 1)^2

This is a repeated linear factor problem. A repeated factor needs one term per power: A/(x + 1) plus B/(x + 1)².

More repeated linear factor examples

x²/(x − 1)³ A cubed factor contributes three terms, one for each power. 8/(x³ + 2x² + x) Factoring gives x(x + 1)², a mix of a simple and a repeated root.

All worked examples

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

Frequently asked questions

How does 1/(x + 1)² decompose? ▾

1/(x + 1)² = 1/(x + 1)^2. A repeated factor needs one term per power: A/(x + 1) plus B/(x + 1)².

How was the denominator factored? ▾

It factors as (x + 1)^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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