Partial Fractions with Repeated Factors

students, in this lesson you will learn how repeated factors work in partial fraction decomposition and why they matter for integrating rational functions 📘. Repeated factors show up when the denominator of a rational function contains the same factor more than once, such as $(x-2)^3$ or $(x^2+1)^2$. By the end of this lesson, you should be able to explain the idea, set up the correct decomposition, and use it to integrate rational expressions accurately.

What repeated factors mean

A rational function is a quotient of polynomials, written as $\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials and $Q(x)\neq 0$. In partial fractions, we rewrite a rational function as a sum of simpler fractions. This is especially useful for integration because simpler fractions are easier to integrate.

A factor is repeated when it appears more than once in the denominator. For example, in $\frac{5}{(x-1)^2}$, the factor $(x-1)$ is repeated twice. In $\frac{3x+1}{(x+2)^4(x^2+1)}$, the factor $(x+2)$ is repeated four times.

The key idea is that when a factor repeats, the decomposition must include a term for every power of that factor, starting with the first power and ending with the highest power. This rule is not arbitrary; it ensures the partial fraction form has enough flexibility to match the original rational expression.

For a repeated linear factor $(x-a)^n$, the decomposition includes

$$\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_n}{(x-a)^n}.$$

For a repeated irreducible quadratic factor $(x^2+bx+c)^n$, the decomposition includes numerators that are linear, because the quadratic cannot be factored further over the real numbers:

$$\frac{A_1x+B_1}{x^2+bx+c}+\frac{A_2x+B_2}{(x^2+bx+c)^2}+\cdots+\frac{A_nx+B_n}{(x^2+bx+c)^n}.$$

This pattern is a major part of partial fraction decomposition in Calculus 2.

Why repeated factors need multiple terms

students, imagine you are trying to describe a staircase with only one step. You would miss the higher steps. Repeated factors work the same way: each power contributes a different piece, so one fraction alone is not enough.

Suppose the denominator has $(x-3)^3$. If you only wrote $\frac{A}{x-3}$, you would not have enough degrees of freedom to match every possible numerator that could arise after clearing denominators. The terms

$$\frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{C}{(x-3)^3}$$

allow the decomposition to capture all possible behavior near $x=3$.

This is especially important when integrating. For instance, $\int \frac{1}{(x-3)^3}\,dx$ is very different from $\int \frac{1}{x-3}\,dx$. The first becomes a power rule problem after rewriting, while the second becomes a logarithm. Since repeated factors can create both kinds of terms, partial fractions separates them so each piece can be handled correctly.

A useful fact is that repeated factors often produce a mix of logarithms and powers during integration. The lowest power term, like $\frac{A}{x-a}$, integrates to a logarithm. Higher powers, like $\frac{B}{(x-a)^2}$ or $\frac{C}{(x-a)^3}$, integrate using the power rule after rewriting as $(x-a)^{-2}$ or $(x-a)^{-3}$.

Repeated linear factors: setup and examples

Let’s look at repeated linear factors first. Suppose you need to decompose

$$\frac{7x+5}{(x-1)^2(x+4)}.$$

Because $(x-1)$ is repeated twice, the decomposition must include both powers of that factor:

$$\frac{7x+5}{(x-1)^2(x+4)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+4}.$$

To find $A$, $B$, and $C$, multiply both sides by $(x-1)^2(x+4)$:

$$7x+5=A(x-1)(x+4)+B(x+4)+C(x-1)^2.$$

Then expand and match coefficients or plug in convenient values of $x$.

For example, if $x=1$:

$$7(1)+5=B(1+4),$$

$$12=5B,$$

$$B=\frac{12}{5}.$$

If $x=-4$:

$$7(-4)+5=C(-5)^2,$$

$$-23=25C,$$

$$C=-\frac{23}{25}.$$

Then use another value, such as $x=0$, to find $A$:

$$5=A(-1)(4)+B(4)+C(1),$$

which gives a linear equation in $A$.

This example shows a common strategy: repeated linear factors often make it convenient to substitute values that zero out many terms. That makes the system easier to solve 😊.

Now consider an integration example:

$$\int \frac{7x+5}{(x-1)^2(x+4)}\,dx.$$

After decomposition, you would integrate

$$\int \left(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+4}\right)dx.$$

This becomes

$$A\ln|x-1|-\frac{B}{x-1}+C\ln|x+4|+K,$$

because

$$\int \frac{1}{x-a}\,dx=\ln|x-a|+K$$

and

$$\int (x-a)^{-2}\,dx=-(x-a)^{-1}+K.$$

Repeated irreducible quadratic factors

Repeated factors are not limited to linear terms. Sometimes the repeated factor is an irreducible quadratic, meaning it cannot be factored further using real numbers. Examples include $x^2+1$ and $x^2+4x+5$.

If the denominator contains $(x^2+1)^2$, the decomposition must include two terms:

$$\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}.$$

Each numerator must be linear, not constant. That is because when you combine the fractions, the numerator must be flexible enough to match any polynomial of degree less than $2$ over that factor.

For example, decompose

$$\frac{3x^2+2x+1}{(x^2+1)^2}.$$

A correct form is

$$\frac{3x^2+2x+1}{(x^2+1)^2}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}.$$

Multiply both sides by $(x^2+1)^2$:

$$3x^2+2x+1=(Ax+B)(x^2+1)+(Cx+D).$$

After expanding,

$$3x^2+2x+1=Ax^3+Bx^2+Ax+B+Cx+D.$$

Now compare coefficients. Since the left side has no $x^3$ term, $A=0$. Then match the remaining coefficients. This process shows why the form must include both factors and why the numerators have the shapes they do.

When integrating expressions with repeated irreducible quadratics, the result may include arctangent terms. For example,

$$\int \frac{1}{x^2+1}\,dx=\arctan(x)+K,$$

while higher powers, such as $\int \frac{1}{(x^2+1)^2}\,dx$, require more advanced methods after decomposition, often involving algebraic manipulation and standard integrals.

How repeated factors fit into the full partial fractions process

Repeated factors are one part of the larger partial fractions method. The overall process usually follows these steps:

Make sure the rational function is proper, meaning the degree of the numerator is less than the degree of the denominator. If not, use polynomial long division first.
Factor the denominator completely over the real numbers.
Write the correct partial fraction form.
Solve for the unknown constants.
Integrate each term separately if the goal is integration.

Repeated factors affect step 3 the most. They determine how many terms appear for a given factor. For a factor repeated $n$ times, you need $n$ terms. This rule applies whether the factor is linear or irreducible quadratic.

The reason this matters in Calculus 2 is that many integration problems are designed so the original integrand looks complicated, but after decomposition, it becomes a sum of standard integrals. Repeated factors are common in these problems because they test whether you know the correct form before solving.

For example, if you see

$$\frac{P(x)}{(x-2)^3(x^2+1)^2},$$

you should immediately write

$$\frac{P(x)}{(x-2)^3(x^2+1)^2}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}.$$

That setup is the heart of repeated-factor partial fractions.

Common mistakes to avoid

One common mistake is skipping powers. If the denominator has $(x-5)^4$, you need all four terms:

$$\frac{A}{x-5}+\frac{B}{(x-5)^2}+\frac{C}{(x-5)^3}+\frac{D}{(x-5)^4}.$$

Another mistake is using constants where linear numerators are required for irreducible quadratics. For $(x^2+1)^2$, the correct numerator pattern is linear for each term, not constant.

A third mistake is forgetting to factor out the repeated structure before integrating. If you try to integrate directly without decomposition, you may miss the standard forms that make the problem manageable.

Finally, always check that your decomposition makes sense by combining terms mentally or by matching the degree pattern. This helps catch algebra errors before they affect the integral.

Conclusion

Repeated factors are an essential part of partial fractions because they tell you how many terms must appear for each factor in the denominator. students, when you see a factor repeated $n$ times, you must include a term for every power from $1$ through $n$. This rule works for both repeated linear factors and repeated irreducible quadratic factors. Once the correct form is written, the rest of the method follows the same general path: solve for constants, then integrate each simpler fraction using familiar Calculus 2 techniques. Understanding repeated factors makes partial fraction decomposition more accurate, more efficient, and much easier to use in rational function integration ✨.

Study Notes

A repeated factor is a factor that appears more than once in the denominator.
For a repeated linear factor $(x-a)^n$, the partial fraction form includes

$$\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_n}{(x-a)^n}.$$

For a repeated irreducible quadratic factor $(x^2+bx+c)^n$, the form includes linear numerators:

$$\frac{A_1x+B_1}{x^2+bx+c}+\cdots+\frac{A_nx+B_n}{(x^2+bx+c)^n}.$$

Repeated factors require one term for each power from $1$ to $n$.
In integration, $\frac{1}{x-a}$ gives a logarithm, while higher powers like $(x-a)^{-2}$ use the power rule.
Irreducible quadratic terms often lead to arctangent functions after integration.
Always factor the denominator completely before writing the partial fraction decomposition.
Repeated factors are a key part of the broader partial fractions method used to integrate rational functions.