3.1 - Partial fractions

3.1 - Partial fractions

Adding fractions creates a single rational expression; partial fractions reverse that process. In this part of 3.1, you will split fractions with distinct linear factors, a repeated factor or an irreducible quadratic. Choosing the right numerator forms will turn each decomposition into ordinary algebra and prepare expressions for later integration.

Distinct linear factors

Can 5x+1(x−1)(x+2)\dfrac{5x+1}{(x-1)(x+2)} be written as two simpler fractions? Adding 2/(x−1)2/(x-1) and 3/(x+2)3/(x+2) gives exactly this numerator. The challenge is to find those constants when they are unknown.

Partial fractions

A partial-fraction decomposition writes a rational function as a sum of simpler rational functions. A fraction is proper when its numerator has smaller degree than its denominator. The decomposition has the same values as the original wherever the original denominator is nonzero.

For two distinct linear factors, start with

N(x)(ax+b)(cx+d)=Aax+b+Bcx+d.\frac{N(x)}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d}.

For three distinct linear factors, include one constant numerator over each factor. The factors need not have leading coefficient 11. These forms are not in the formula booklet: learn how to choose them.

Multiply through by the whole denominator. This gives a polynomial identity: equal coefficients on both sides, or equal values for every input. Substituting a factor's zero then removes all but one unknown. The original fraction is undefined at that input, but the cleared polynomial identity is valid there; this is why the substitution works.

Worked example 1: two distinct factors

Question. Express 5x+1(x−1)(x+2)\dfrac{5x+1}{(x-1)(x+2)} in partial fractions.

Plan. Put one constant above each factor, clear the denominator and use the two factor zeros.

5x+1(x−1)(x+2)=Ax−1+Bx+2⟹5x+1=A(x+2)+B(x−1).\frac{5x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2} \quad\Longrightarrow\quad 5x+1=A(x+2)+B(x-1).

At x=1x=1, 6=3A6=3A, so A=2A=2. At x=−2x=-2, −9=−3B-9=-3B, so B=3B=3.

Answer. 2x−1+3x+2\dfrac2{x-1}+\dfrac3{x+2}, for x≠1,−2x\ne1,-2.

Check. Recombining gives numerator 2(x+2)+3(x−1)=5x+12(x+2)+3(x-1)=5x+1, verifying the identity for every allowed input.

Keep the numerator and add a third, non-monic factor to the denominator. Its zero is −1/2-1/2, not −1-1.

Worked example 2: a third linear factor

Question. Express 5x+1(x−1)(x+2)(2x+1)\dfrac{5x+1}{(x-1)(x+2)(2x+1)} in partial fractions.

Plan. Use three constants and substitute the three distinct zeros in the cleared identity.

5x+1=A(x+2)(2x+1)+B(x−1)(2x+1)+C(x−1)(x+2).5x+1=A(x+2)(2x+1)+B(x-1)(2x+1)+C(x-1)(x+2).

The useful inputs give

x=1:6=9A⟹A=23x=−2:−9=9B⟹B=−1x=−12:−32=−94C⟹C=23\begin{aligned} x=1:&\quad6=9A &&\Longrightarrow A=\frac23\\ x=-2:&\quad-9=9B &&\Longrightarrow B=-1\\ x=-\frac12:&\quad-\frac32=-\frac94C &&\Longrightarrow C=\frac23 \end{aligned}

Answer. 23(x−1)−1x+2+23(2x+1)\dfrac{2}{3(x-1)}-\dfrac1{x+2}+\dfrac{2}{3(2x+1)}, excluding the three denominator zeros.

Check. Expanding the cleared right side gives

23(2x2+5x+2)−(2x2−x−1)+23(x2+x−2)=5x+1.\frac23(2x^2+5x+2)-(2x^2-x-1)+\frac23(x^2+x-2)=5x+1.

The squared terms cancel, as they must.

You can also expand the identity and equate coefficients. This gives simultaneous linear equations and is especially useful when substitution leaves an unknown unresolved.

Watch out

Do not split a numerator term by term over separate factors. In general, (u+v)/(pq)≠u/p+v/q(u+v)/(pq)\ne u/p+v/q. Choose unknown numerators, then verify by recombining.

Use all three factor zeros in a fresh decomposition.

Repeated factors

A repeated square needs two fractions, one over the factor and one over its square. For example,

N(x)(x−1)(x+2)2=Ax−1+Bx+2+C(x+2)2.\frac{N(x)}{(x-1)(x+2)^2} =\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}.

The BB term is essential: after clearing the denominator, it contributes a different polynomial from the CC term. Omitting it generally leaves too few adjustable coefficients to match the numerator.

A repeated linear factor

For a proper fraction with distinct linear factors ax+bax+b and cx+dcx+d, one of them squared, use

N(x)(ax+b)(cx+d)2=Aax+b+Bcx+d+C(cx+d)2.\frac{N(x)}{(ax+b)(cx+d)^2} =\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{(cx+d)^2}.

Include every power up to the repeated square. Not in the formula booklet: learn the form.

Substituting the repeated factor's zero finds the constant over its square. Find the remaining constant using another convenient input or a coefficient comparison.

Worked example 3: the second factor is squared

Question. Express 5x+1(x−1)(x+2)2\dfrac{5x+1}{(x-1)(x+2)^2} in partial fractions.

Plan. Include both powers of x+2x+2, clear the denominator and finish with a coefficient comparison.

5x+1=A(x+2)2+B(x−1)(x+2)+C(x−1).5x+1=A(x+2)^2+B(x-1)(x+2)+C(x-1).

At x=1x=1, 6=9A6=9A, giving A=2/3A=2/3. At x=−2x=-2, −9=−3C-9=-3C, giving C=3C=3. Comparing squared coefficients gives 0=A+B0=A+B, so B=−2/3B=-2/3.

Answer. 23(x−1)−23(x+2)+3(x+2)2\dfrac{2}{3(x-1)}-\dfrac{2}{3(x+2)}+\dfrac3{(x+2)^2}, for x≠1,−2x\ne1,-2.

Check. The combined numerator is

23(x+2)2−23(x−1)(x+2)+3(x−1)=2(x+2)+3(x−1)=5x+1.\frac23(x+2)^2-\frac23(x-1)(x+2)+3(x-1) =2(x+2)+3(x-1)=5x+1.

Now square the other factor, keeping the numerator and the two zeros. The squared-denominator term must move with its factor.

Continue this lesson.

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