P4 6.3 - Integrating rational functions

P4 6.3 - Integrating rational functions

Use partial fractions to turn rational functions into logarithms and powers that you can integrate. Distinguish simple from repeated factors, divide improper fractions first, and check the domain before evaluating a definite integral.

Using distinct-factor partial fractions

How can a quotient with two denominator factors produce two logarithms? Split it into the simpler fractions before choosing an integration rule.

You already know how to find partial fractions by clearing the denominator. Integration is then term by term. A linear denominator gives a logarithm, with a factor compensating for its derivative.

Integrating a linear reciprocal

For a≠0a\ne0, on an interval where ax+b≠0ax+b\ne0,

∫Kax+b dx=Kaln⁡∣ax+b∣+C.\int\frac{K}{ax+b}\,dx=\frac Ka\ln|ax+b|+C.

The absolute value keeps the logarithm real on either side of the excluded zero; it does not permit integration across that zero.

Not in the formula booklet: learn it.

Check the fractions before integrating, then check the logarithm coefficients separately.

Worked example 1: two logarithms

Find ∫5x+7(x+1)(x+2) dx\int\frac{5x+7}{(x+1)(x+2)}\,dx.

Plan. Decompose using the two distinct linear factors, then integrate each fraction.

5x+7=A(x+2)+B(x+1).5x+7=A(x+2)+B(x+1).

Putting x=−1x=-1 gives A=2A=2; putting x=−2x=-2 gives B=3B=3. Hence

∫(2x+1+3x+2)dx=2ln⁡∣x+1∣+3ln⁡∣x+2∣+C.\int\left(\frac2{x+1}+\frac3{x+2}\right)dx =2\ln|x+1|+3\ln|x+2|+C.

Answer. 2ln⁡∣x+1∣+3ln⁡∣x+2∣+C2\ln|x+1|+3\ln|x+2|+C, on any interval avoiding −1,−2-1,-2.

Check. Differentiating and recombining gives numerator 2(x+2)+3(x+1)=5x+72(x+2)+3(x+1)=5x+7, exactly the original integrand.

Change only x+1x+1 to 2x+12x+1 in the denominator. Recalculate the fractions and compensate for the new inner derivative 22.

Worked example 2: a scaled linear factor

Find ∫5x+7(2x+1)(x+2) dx\int\frac{5x+7}{(2x+1)(x+2)}\,dx.

Plan. Find A,BA,B in A/(2x+1)+B/(x+2)A/(2x+1)+B/(x+2), then divide the first logarithm coefficient by 22.

5x+7=A(x+2)+B(2x+1).5x+7=A(x+2)+B(2x+1).

At x=−1/2x=-1/2, 9/2=3A/29/2=3A/2, so A=3A=3. At x=−2x=-2, −3=−3B-3=-3B, so B=1B=1. Therefore

∫(32x+1+1x+2)dx=32ln⁡∣2x+1∣+ln⁡∣x+2∣+C.\int\left(\frac3{2x+1}+\frac1{x+2}\right)dx =\frac32\ln|2x+1|+\ln|x+2|+C.

Answer. 32ln⁡∣2x+1∣+ln⁡∣x+2∣+C\frac32\ln|2x+1|+\ln|x+2|+C, avoiding −1/2,−2-1/2,-2.

Check. Its derivative recombines to numerator 3(x+2)+(2x+1)=5x+73(x+2)+(2x+1)=5x+7.

The opening quotient produces two logarithms because its decomposition has two linear reciprocals. Do not try to take the logarithm of a whole denominator unless its derivative is proportional to the numerator.

Watch out

∫3/(2x+1) dx=32ln⁡∣2x+1∣+C\int3/(2x+1)\,dx=\frac32\ln|2x+1|+C, not 3ln⁡∣2x+1∣+C3\ln|2x+1|+C. Differentiate to catch the missing factor.

Change only the numerator of Worked example 2.

Repeated-factor integrals

A denominator squared produces a negative power to integrate, not a second logarithm. Use the power rule except when the integrand power is −1-1.

For example, ∫3(x−1)−2 dx=−3(x−1)−1+C\int3(x-1)^{-2}\,dx=-3(x-1)^{-1}+C. More generally, for n≠1n\ne1,

∫K(ax+b)n dx=Ka(1−n)(ax+b)1−n+C,\int\frac{K}{(ax+b)^n}\,dx =\frac{K}{a(1-n)}(ax+b)^{1-n}+C,

where a≠0a\ne0 and the denominator is nonzero. This is the power rule and reverse chain rule, both already familiar.

Worked example 3: separate powers of a repeated factor

Find ∫3x+1(x−1)2 dx\int\frac{3x+1}{(x-1)^2}\,dx.

Plan. Include both 1/(x−1)1/(x-1) and 1/(x−1)21/(x-1)^2, then choose logarithm or power for each.

The numerator identity 3x+1=3(x−1)+43x+1=3(x-1)+4 gives

∫3x+1(x−1)2 dx=∫(3x−1+4(x−1)−2)dx=3ln⁡∣x−1∣−4x−1+C.\begin{aligned} \int\frac{3x+1}{(x-1)^2}\,dx &=\int\left(\frac3{x-1}+4(x-1)^{-2}\right)dx\\ &=3\ln|x-1|-\frac4{x-1}+C. \end{aligned}

Answer. 3ln⁡∣x−1∣−4/(x−1)+C3\ln|x-1|-4/(x-1)+C, for x≠1x\ne1.

Check. The derivative has numerator 3(x−1)+4=3x+13(x-1)+4=3x+1 over (x−1)2(x-1)^2.

Change only the denominator to (2x−1)2(2x-1)^2. Both resulting terms need the inner-derivative factor.

Worked example 4: repeated scaled factor

Find ∫3x+1(2x−1)2 dx\int\frac{3x+1}{(2x-1)^2}\,dx.

Plan. Write 3x+1=32(2x−1)+523x+1=\frac32(2x-1)+\frac52, then integrate both pieces.

∫3x+1(2x−1)2 dx=∫(3/22x−1+5/2(2x−1)2)dx=34ln⁡∣2x−1∣−54(2x−1)+C.\begin{aligned} \int\frac{3x+1}{(2x-1)^2}\,dx &=\int\left(\frac{3/2}{2x-1}+\frac{5/2}{(2x-1)^2}\right)dx\\ &=\frac34\ln|2x-1|-\frac5{4(2x-1)}+C. \end{aligned}

Answer. 34ln⁡∣2x−1∣−5/[4(2x−1)]+C\frac34\ln|2x-1|-5/[4(2x-1)]+C, avoiding x=1/2x=1/2.

Check. Differentiation gives 3/22x−1+5/2(2x−1)2\frac{3/2}{2x-1}+\frac{5/2}{(2x-1)^2}. Recombining yields numerator 32(2x−1)+52=3x+1\frac32(2x-1)+\frac52=3x+1.

With an additional distinct factor, decompose first into all required pieces, such as A/(x−1)+B/(x+2)+D/(x+2)2A/(x-1)+B/(x+2)+D/(x+2)^2, then integrate the first two as logarithms and the last as a power. Not every rational function needs decomposition: ∫x/(x2+5) dx=12ln⁡(x2+5)+C\int x/(x^2+5)\,dx=\frac12\ln(x^2+5)+C follows directly from a denominator derivative. Similarly,

∫2(2x−1)4 dx=−13(2x−1)3+C.\int\frac2{(2x-1)^4}\,dx=-\frac1{3(2x-1)^3}+C.

Differentiating the last expression returns 2(2x−1)−42(2x-1)^{-4}.

Watch out

∫(x−1)−2 dx\int(x-1)^{-2}\,dx is −(x−1)−1+C-(x-1)^{-1}+C. A logarithm is for the power −1-1; it is not a rule for every fraction.

Change only the numerator coefficient in the scaled-factor example.

Improper fractions before integration

An improper algebraic fraction has numerator degree at least as large as denominator degree. Divide first so that its polynomial part is not lost.

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