Partial Fractions: Learn It 3

Giselle Miller

Partial Fractions: Learn It 3

Repeated Linear Factors

Sometimes you’ll encounter denominators where the same linear factor appears multiple times. This creates a different pattern for your partial fraction decomposition.

repeated linear factors

If your denominator contains a repeated linear factor [latex]{\left(ax+b\right)}^{n}[/latex] where [latex]n \geq 2[/latex], then your partial fraction decomposition must include all powers from [latex]1[/latex] up to [latex]n[/latex]:

[latex]\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\cdots +\frac{A_n}{(ax+b)^n}[/latex]

Each power of the repeated factor contributes to the overall decomposition—you can’t skip any of them. Let’s look at what this pattern looks like in practice:

If your denominator has [latex](x-2)^3[/latex], then you need: [latex]\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}[/latex]

If your denominator has [latex](3x+1)^2[/latex], then you need: [latex]\frac{A}{3x+1}+\frac{B}{(3x+1)^2}

Warning! Don't forget the lower powers! A common mistake is to only include the highest power [latex](ax+b)^n[/latex] and miss the terms with[latex](ax+b)^1, (ax+b)^2[/latex], etc.

The technique for finding the constants follows the same approach as before, but you'll work through more algebra since there are more unknowns to solve for. Let's see how this plays out in the next example.

Evaluate [latex]\displaystyle\int \frac{x - 2}{{\left(2x - 1\right)}^{2}\left(x - 1\right)}dx[/latex].

We have [latex]\text{degree}\left(x - 2\right)<\text{degree}\left({\left(2x - 1\right)}^{2}\left(x - 1\right)\right)[/latex], so we can proceed with the decomposition. Since [latex]{\left(2x - 1\right)}^{2}[/latex] is a repeated linear factor, include [latex]\frac{A}{2x - 1}+\frac{B}{{\left(2x - 1\right)}^{2}}[/latex] in the decomposition. Thus,

[latex]\frac{x - 2}{{\left(2x - 1\right)}^{2}\left(x - 1\right)}=\frac{A}{2x - 1}+\frac{B}{{\left(2x - 1\right)}^{2}}+\frac{C}{x - 1}[/latex].

After getting a common denominator and equating the numerators, we have

[latex]x - 2=A\left(2x - 1\right)\left(x - 1\right)+B\left(x - 1\right)+C{\left(2x - 1\right)}^{2}[/latex].

We then use the method of equating coefficients to find the values of [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex].

[latex]x - 2=\left(2A+4C\right){x}^{2}+\left(-3A+B - 4C\right)x+\left(A-B+C\right)[/latex].

Equating coefficients yields [latex]2A+4C=0[/latex], [latex]-3A+B - 4C=1[/latex], and [latex]A-B+C=-2[/latex]. Solving this system yields [latex]A=2[/latex], [latex]B=3[/latex], and [latex]C=-1[/latex].

Alternatively, we can use the method of strategic substitution. In this case, substituting [latex]x=1[/latex] and [latex]x=\frac{1}{2}[/latex] into the previous equation easily produces the values [latex]B=3[/latex] and [latex]C=-1[/latex]. At this point, it may seem that we have run out of good choices for [latex]x[/latex], however, since we already have values for [latex]B[/latex] and [latex]C[/latex], we can substitute in these values and choose any value for [latex]x[/latex] not previously used. The value [latex]x=0[/latex] is a good option. In this case, we obtain the equation [latex]-2=A\left(-1\right)\left(-1\right)+3\left(-1\right)+\left(-1\right){\left(-1\right)}^{2}[/latex] or, equivalently, [latex]A=2[/latex].

Now that we have the values for [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex], we rewrite the original integral and evaluate it:

[latex]\begin{array}{cc}\hfill {\displaystyle\int \frac{x - 2}{{\left(2x - 1\right)}^{2}\left(x - 1\right)}dx}& ={\displaystyle\int \left(\frac{2}{2x - 1}+\frac{3}{{\left(2x - 1\right)}^{2}}-\frac{1}{x - 1}\right)dx}\hfill \\ & =\text{ln}|2x - 1|-\frac{3}{2\left(2x - 1\right)}-\text{ln}|x - 1|+C.\end{array}[/latex]