H6PureStretch

Integration Using Partial Fractions

A rational function that resists direct integration can be split into partial fractions, each of which integrates to a logarithm. This connects algebra and calculus neatly.

30 min Integration

The big picture

Decompose before integrating. Distinct linear denominators produce logarithms, but a squared linear denominator is a negative power and usually integrates using the power rule. Do not turn every partial fraction into a logarithm. Keep excluded x-values from the original denominator. For a definite integral, check that no singularity lies inside the interval before evaluating an antiderivative at the endpoints. A coefficient inside a linear denominator also changes the integral: $\int A/(2x+1)\,dx=(A/2)\ln|2x+1|+C$ . Keeping the factors in their original form helps you see that coefficient. For a squared factor, integrate it as a power and apply the same reverse-chain adjustment.

Edexcel A level Maths: 11.7 Integrating Using Partial FractionsCurated by Kepler, not produced by us. Watch it, then work through the notes below.

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What you'll be able to do

Split a rational function into partial fractions
Integrate each term to a logarithm
Combine logs with the laws of logarithms
Evaluate definite integrals

The method

Decompose into partial fractions, then integrate each $\dfrac{A}{x + a}$ term to $A\ln|x + a|$ . Combine the logarithms at the end.

\int \frac{A}{x+a}\,dx = A\ln|x+a| + c

Each partial fraction integrates to a log.

Example 1Find

\int \dfrac{1}{x(x+1)}\,dx

1Partial fractions:

\tfrac{1}{x} - \tfrac{1}{x+1}

\ln|x| - \ln|x+1| + c = \ln\left|\tfrac{x}{x+1}\right| + c

Answer

\ln\left|\dfrac{x}{x+1}\right| + c

Tip — Combine the logs with log laws to give a tidy single-log answer.

Repeated factors produce a power term as well as logs

Keep excluded x-values from the original denominator. For a definite integral, check that no singularity lies inside the interval before evaluating an antiderivative at the endpoints.

A coefficient inside a linear denominator also changes the integral: $\int A/(2x+1)\,dx=(A/2)\ln|2x+1|+C$ . Keeping the factors in their original form helps you see that coefficient. For a squared factor, integrate it as a power and apply the same reverse-chain adjustment.

Example 1Integrate

\dfrac{3x+1}{(x-1)(x+1)^2}

1Write

A/(x-1)+B/(x+1)+C/(x+1)^2

. Clearing denominators gives

3x+1=A(x+1)^2+B(x-1)(x+1)+C(x-1)

2At

x=1

A=1

; at

x=-1

C=1

; comparing

x^2

terms gives

B=-1

3Integrate

1/(x-1)-1/(x+1)+1/(x+1)^2

termwise.

4The squared term integrates to

-1/(x+1)

, not a logarithm.

Answer

\ln|x-1|-\ln|x+1|-\dfrac1{x+1}+C

, on intervals excluding

x=\pm1

Think like an examiner

Write the defining equations and show the substitution before evaluating.
Check the domain, units and interpretation of the final result.

Common misconceptions

Integrating the rational function directly.

Split into partial fractions first.

Dropping the modulus in the logarithm.

It is ln|x + a|.

Key results to remember

\int \frac{A}{x+a}\,dx = A\ln|x+a| + c

Standard log integral.

\ln p - \ln q = \ln\tfrac{p}{q}

Combine logs.

Key takeaways

Decompose into partial fractions.
Each A/(x+a) integrates to A ln|x+a|.
Combine logs into a single tidy logarithm.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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