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Rational expressions

Partial Fraction Decomposition: Setup, Coefficients and Checks

Partial fraction decomposition rewrites a rational expression as simpler fractions. The main decision is the denominator structure: distinct factors, repeated powers and irreducible quadratics require different numerator templates. Decomposition is an algebra step, not integration itself.

Direct answer

3x+5(x+1)(x+2)=2x+1+1x+2\frac{3x+5}{(x+1)(x+2)}=\frac{2}{x+1}+\frac{1}{x+2}

Keep every original denominator restriction; here x ≠ −1,−2. The calculator handles supported rational expressions, not every possible factorization.

Partial fraction decomposition steps before integration

Compare polynomial degrees first. If the numerator degree is at least the denominator degree, perform polynomial division and decompose only the proper remainder fraction.

Factor the denominator and record excluded roots. For (x+1)(x+2), use A/(x+1)+B/(x+2). A repeated factor (x−r)ᵏ needs a separate term for every power from 1 to k.

Multiply the identity by the common denominator. For this example, 3x+5=A(x+2)+B(x+1). Matching coefficients gives A+B=3 and 2A+B=5, so A=2 and B=1.

An irreducible quadratic such as x²+1 needs a numerator Ax+B rather than just A. This is the mathematical template; do not assume all such cases are implemented by the calculator.

Recombine the fractions over a common denominator and compare the full numerator with the original. Solving coefficients is not enough if a term was omitted from the template.

1
3 x+5=A (x+2)+B (x+1)3\,x + 5 = A\,\left(x + 2\right) + B\,\left(x + 1\right)
2
A+B=3; 2A+B=5
3
A=2; B=1

Worked examples

3 x+5(x+1) (x+2)\frac{3\,x + 5}{\left(x + 1\right)\,\left(x + 2\right)}
2x+1+1x+2\frac{2}{x + 1} + \frac{1}{x + 2}

Distinct linear factors; exclude −1 and −2.

1x (x+1)\frac{1}{x\,\left(x + 1\right)}
1x−1x+1\frac{1}{x} - \frac{1}{x + 1}

The numerators recombine to 1; exclude 0 and −1.

x+2x2 (x+1)\frac{x + 2}{x^{2}\,\left(x + 1\right)}
−1x+2x2+1x+1\frac{-1}{x} + \frac{2}{x^{2}} + \frac{1}{x + 1}

The repeated x factor requires both 1/x and 1/x² terms; exclude 0 and −1.

Common mistakes

  • • Decomposing an improper fraction before polynomial division.
  • • Omitting lower powers of a repeated factor.
  • • Using a constant numerator for an irreducible quadratic.
  • • Dropping excluded roots after a cancellation.
  • • Integrating before checking the decomposition.

Try the worked examples in the calculator

Each link opens the matching calculator with the expression, variable and required conditions already filled in.

The button opens the decomposition tool, not the integral solver. Check the decomposition before using the resulting terms in an integration problem.

部分分数分解計算機:3 x+5x2+3 x+2\frac{3\,x + 5}{x^{2} + 3\,x + 2}部分分数分解計算機:1x (x+1)\frac{1}{x\,\left(x + 1\right)}部分分数分解計算機:x+2x2 (x+1)\frac{x + 2}{x^{2}\,\left(x + 1\right)}
Open the calculator without a preset: 部分分数分解計算機

References

Frequently asked questions

Why do repeated factors need several terms?+

A denominator (x−r)ᵏ requires terms for (x−r), (x−r)² through (x−r)ᵏ. Omitting one can make coefficient matching impossible.

How do I check the coefficients?+

Multiply by the original denominator and compare polynomial coefficients, or recombine all fractions into one expression. Keep the original domain restrictions.

Does decomposition give the integral?+

No. It gives an equivalent rational expression on its domain. Each resulting term must still be integrated with the appropriate rule.

Conclusion

For partial fraction decomposition, check degrees, choose the full factor template, solve coefficients and recombine. Only then use the simpler fractions in a larger calculation.

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