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Calculus learning

Integrating Rational Functions: Division, Partial Fractions and Domains

Integrating rational functions means integrating a quotient of polynomials. The degree test determines the first step: a numerator at least as large as the denominator needs polynomial division before partial fractions. Factoring the denominator then determines the terms to use.

Direct answer

Divide first if needed, decompose the proper fraction, then integrate its simpler pieces.

The denominator must not be zero. An antiderivative applies separately on each interval of the original domain.

Integrating Rational Functions worked method

For (3x+5)/(x²+3x+2), factor the denominator as (x+1)(x+2). Neither x=−1 nor x=−2 belongs to the domain.

Write A/(x+1)+B/(x+2), multiply by the original denominator, and compare coefficients: A+B=3 and 2A+B=5. Hence A=2, B=1.

Integrate to 2ln|x+1|+ln|x+2|+C. Differentiate the answer and combine its fractions to check the original integrand.

Repeated linear factors require every power up to the multiplicity. An irreducible quadratic gets a linear numerator, not just a constant.

An irreducible quadratic may lead to arctan after completing the square. Keep its scale factor; for 1/(x²+a²), a positive constant a gives arctan(x/a)/a.

1
3 x+5(x+1) (x+2)=2x+1+1x+2\frac{3\,x + 5}{\left(x + 1\right)\,\left(x + 2\right)} = \frac{2}{x + 1} + \frac{1}{x + 2}
2
∫3 x+5x2+3 x+2 dx\int \frac{3\,x + 5}{x^{2} + 3\,x + 2}\,\mathrm{d}x
3
∫x2+1x+1 dx\int \frac{x^{2} + 1}{x + 1}\,\mathrm{d}x

Worked examples

∫1x (x+1) dx\int \frac{1}{x\,\left(x + 1\right)}\,\mathrm{d}x
ln⁡∣x∣−ln⁡∣x+1∣+C\ln\left|x\right| - \ln\left|x + 1\right| + C

Exclude 0 and −1.

∫1x2+4 dx\int \frac{1}{x^{2} + 4}\,\mathrm{d}x
arctan⁡(x2)2+C\frac{\arctan\left(\frac{x}{2}\right)}{2} + C

The factor 1/2 comes from the inner derivative.

∫x2+1x+1 dx\int \frac{x^{2} + 1}{x + 1}\,\mathrm{d}x
x²/2−x+2ln|x+1|+C

Division gives x−1+2/(x+1).

Common mistakes

  • • Skipping division for an improper fraction.
  • • Omitting intermediate powers of a repeated factor.
  • • Dropping absolute values from real logarithmic antiderivatives.

Try the supported calculator preset

The preset opens a supported expression with its variable and conditions. Teaching examples elsewhere in this guide are reference explanations, not claims that the engine supports every method.

The direct original rational integral is currently unsupported. First decompose (3*x+5)/(x^2+3*x+2) in the partial-fraction tool; this preset then integrates the equivalent expression 2/(x+1)+1/(x+2). Excluded points remain −1 and −2.

積分:2x+1+1x+2\frac{2}{x + 1} + \frac{1}{x + 2}
Open the calculator without a preset: 積分

References

Frequently asked questions

Is decomposition the same as integration?+

No. Decomposition is an algebraic identity; integrate each resulting term afterward.

Can I cancel a denominator and restore excluded points?+

No. Cancelling can simplify the expression but does not change the original domain.

What if the engine cannot factor my polynomial?+

It may return unsupported. That is a limitation of its factoring rules, not of partial fractions as a mathematical method.

Conclusion

Integrating Rational Functions starts with the stated domain and conditions. Recheck those before carrying a worked example into a different problem.

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