1.08j - Integrating rational functions with partial fractions

1.08j - Integrating rational functions with partial fractions

Break a rational integrand into simple linear-factor fractions, then integrate the logarithmic and power terms. Handle repeated factors, linear scaling and valid intervals without losing excluded inputs.

Turning distinct linear factors into logarithms

Can a denominator with three factors produce only two logarithms? The numerator decides: 2x+42x+4 contains a factor that can cancel. Before that surprise, build the basic two-factor method carefully.

You already know

Partial fractions give an identity on the original domain. Clear denominators and substitute the factor roots to find coefficients. If the numerator's degree is at least the denominator's, divide first.

Substituting a pole into the cleared polynomial identity is legal; substituting it into the original rational function is not.

Integrating a simple linear fraction

For a≠0a\ne0, on intervals avoiding ax+b=0ax+b=0,

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

This follows from the supplied logarithmic-derivative formula; it is not separately printed. Not in the formula sheet: learn it.

The sign and size of aa both matter: ∫3/(1−2x) dx=−32ln⁡∣1−2x∣+C\int3/(1-2x)\,\mathrm dx=-\frac32\ln|1-2x|+C.

Worked example 1: two logarithmic terms

Question. Find ∫(3x+5)/[(x+1)(x+2)] dx\int(3x+5)/[(x+1)(x+2)]\,\mathrm dx.

Plan. Decompose first, then integrate each simple fraction.

3x+5=A(x+2)+B(x+1)clear denominatorsx=−1⇒A=2,x=−2⇒B=1I=∫(2x+1+1x+2) dx=2ln⁡∣x+1∣+ln⁡∣x+2∣+C.\begin{aligned} 3x+5&=A(x+2)+B(x+1) &&\text{clear denominators}\\ x=-1&\Rightarrow A=2,\quad x=-2\Rightarrow B=1\\ I&=\int\left(\frac2{x+1}+\frac1{x+2}\right)\,\mathrm dx\\ &=2\ln|x+1|+\ln|x+2|+C. \end{aligned}

Why decompose before integrating? A logarithm of the whole denominator would impose its derivative numerator, which this fraction does not have.

Answer. 2ln⁡∣x+1∣+ln⁡∣x+2∣+C2\ln|x+1|+\ln|x+2|+C, on each interval excluding −2,−1-2,-1.

Check. Differentiating and recombining gives numerator 2(x+2)+(x+1)=3x+52(x+2)+(x+1)=3x+5.

Change only the numerator's constant from 55 to 44. Both coefficients must be recalculated.

Worked example 2: changing the numerator

Question. Find ∫(3x+4)/[(x+1)(x+2)] dx\int(3x+4)/[(x+1)(x+2)]\,\mathrm dx.

Plan. Use the same denominator factors with new constants.

3x+4=A(x+2)+B(x+1)x=−1⇒A=1,x=−2⇒B=2I=ln⁡∣x+1∣+2ln⁡∣x+2∣+C.\begin{aligned} 3x+4&=A(x+2)+B(x+1)\\ x=-1&\Rightarrow A=1,\quad x=-2\Rightarrow B=2\\ I&=\ln|x+1|+2\ln|x+2|+C. \end{aligned}

Why keep modulus? The original rational expression is defined on negative as well as positive intervals away from its poles.

Answer. ln⁡∣x+1∣+2ln⁡∣x+2∣+C\ln|x+1|+2\ln|x+2|+C.

Check. The derivative recombines to ((x+2)+2(x+1))/[(x+1)(x+2)]((x+2)+2(x+1))/[(x+1)(x+2)].

The coefficients determine how much of each logarithmic derivative is present.

Watch out

Wrong: ∫K/(ax+b) dx=Kln⁡∣ax+b∣+C\int K/(ax+b)\,\mathrm dx=K\ln|ax+b|+C for every aa. Right: divide the coefficient by aa, including its sign.

Change the numerator's constant once more.

Including every power of a repeated factor

A squared linear factor contributes two fractions, one for each denominator power. The first integrates to a logarithm; the second integrates by the ordinary power rule.

A repeated linear term

For a≠0a\ne0, away from ax+b=0ax+b=0,

∫K(ax+b)2 dx=−Ka(ax+b)+C.\int\frac{K}{(ax+b)^2}\,\mathrm dx=-\frac{K}{a(ax+b)}+C.

Derived by the power rule with linear scaling. Not in the formula sheet: learn it.

Include both denominator powers in the setup even if a coefficient later turns out to be zero.

Worked example 3: a squared factor

Question. Find ∫(2x+1)/[(x+1)(x−1)2] dx\int(2x+1)/[(x+1)(x-1)^2]\,\mathrm dx.

Plan. Use A/(x+1)+B/(x−1)+D/(x−1)2A/(x+1)+B/(x-1)+D/(x-1)^2.

2x+1=A(x−1)2+B(x+1)(x−1)+D(x+1)x=−1⇒A=−14,x=1⇒D=32A+B=0⇒B=14compare quadratic termsI=−14ln⁡∣x+1∣+14ln⁡∣x−1∣−32(x−1)+C.\begin{aligned} 2x+1&=A(x-1)^2+B(x+1)(x-1)+D(x+1)\\ x=-1&\Rightarrow A=-\frac14,\quad x=1\Rightarrow D=\frac32\\ A+B&=0\Rightarrow B=\frac14 &&\text{compare quadratic terms}\\ I&=-\frac14\ln|x+1|+\frac14\ln|x-1|-\frac{3}{2(x-1)}+C. \end{aligned}

Why the reciprocal term? Its integrand has power −2-2, not −1-1.

Answer. The displayed primitive, on intervals excluding −1,1-1,1.

Check. Differentiating −3/[2(x−1)]-3/[2(x-1)] gives 3/[2(x−1)2]3/[2(x-1)^2], with a positive sign.

Change only the numerator's constant from 11 to 33. The complete denominator structure stays fixed.

Worked example 4: recomputing the repeated-factor coefficients

Question. Find ∫(2x+3)/[(x+1)(x−1)2] dx\int(2x+3)/[(x+1)(x-1)^2]\,\mathrm dx.

Plan. Reuse the full three-term setup, not the previous coefficient values.

x=−1⇒A=14,x=1⇒D=52A+B=0⇒B=−14I=14ln⁡∣x+1∣−14ln⁡∣x−1∣−52(x−1)+C.\begin{aligned} x=-1&\Rightarrow A=\frac14,\quad x=1\Rightarrow D=\frac52\\ A+B&=0\Rightarrow B=-\frac14\\ I&=\frac14\ln|x+1|-\frac14\ln|x-1|-\frac{5}{2(x-1)}+C. \end{aligned}

Why compare the quadratic coefficient? Root substitution finds two coefficients; the identity supplies the third.

Answer. The displayed primitive on the same domain intervals.

Check. Recombination gives numerator (x−1)2/4−(x+1)(x−1)/4+5(x+1)/2=2x+3(x-1)^2/4-(x+1)(x-1)/4+5(x+1)/2=2x+3.

A grouped form with a linear numerator over the squared factor can also be valid, but it must be rewritten to expose both powers before integration. The polynomial identity, rather than a memorised layout alone, checks equivalence.

Watch out

Wrong: retain only A/(x+1)+D/(x−1)2A/(x+1)+D/(x-1)^2. Right: include B/(x−1)B/(x-1) too; omitting it can make the identity impossible.

Use the same complete repeated-factor structure with a new simple factor.

Continue this lesson.

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