Integrating with Partial Fractions

students, imagine you are trying to add up a complicated rate of change, but the expression looks too messy to integrate directly 😅. One very useful IB Mathematics tool is partial fractions. It helps turn a difficult rational function into simpler pieces that are much easier to integrate.

What is partial fraction integration?

A rational function is a fraction of polynomials, such as $\frac{3x+5}{x^2-1}$ or $\frac{2x}{(x+1)(x-2)}$. In calculus, we often want to find $\int f(x)\,dx$ when $f(x)$ is a rational function. If the denominator can be factored, we may be able to rewrite the fraction as a sum of simpler fractions. This process is called partial fraction decomposition.

The main idea is simple: instead of integrating one complicated fraction, we rewrite it as several easier ones. For example,

$$\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}$$

can be rewritten as

$$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$$

for some constants $A$ and $B$. Once those constants are found, each part can usually be integrated using standard rules. This is especially useful because IB calculus expects you to recognise patterns and choose a suitable method 👍.

Why this method matters in calculus

Partial fractions connect directly to the bigger ideas in calculus: simplifying expressions, finding antiderivatives, and using algebra to make integration possible. It is one of the clearest examples of how algebra and calculus work together.

You may see it in:

finding the area under curves when the function is rational,
solving differential equations that produce rational expressions,
modelling real situations where rates involve ratios of polynomials,
simplifying integrals that lead to logarithmic answers.

For IB Mathematics: Analysis and Approaches HL, you should be comfortable recognising when a fraction is suitable for decomposition and when another method might be better. Not every rational function needs partial fractions, but many exam-style questions are designed so that this method is the most efficient one.

Step 1: Make sure the fraction is proper

A rational function is called proper if the degree of the numerator is less than the degree of the denominator. For example, $\frac{2x+1}{x^2+3x+2}$ is proper, but $\frac{x^2+1}{x+1}$ is not.

If the fraction is not proper, you must first use polynomial division.

For example,

$$\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}$$

because

$$x^2+1=(x+1)(x-1)+2$$

Only after this step would you look at the remaining proper fraction. This is important because partial fractions apply to proper rational functions.

Step 2: Factor the denominator

Next, factor the denominator completely if possible. Common factor types include:

distinct linear factors, such as $(x-2)(x+3)$,
repeated linear factors, such as $(x-1)^2$,
irreducible quadratic factors, such as $x^2+1$ over the real numbers.

The factorisation tells you the form of the decomposition.

For example, if

$$\frac{5x+1}{(x-2)(x+1)}$$

then we write

$$\frac{5x+1}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}$$

If the denominator has a repeated factor like $(x-1)^2$, then you must include all powers:

$$\frac{P(x)}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}$$

If the denominator includes a quadratic factor like $x^2+4$, then the numerator over that factor must usually be linear:

$$\frac{P(x)}{x^2+4}=\frac{Ax+B}{x^2+4}$$

This is because the numerator must be one degree lower than the quadratic denominator.

Step 3: Find the constants

After writing the correct decomposition, multiply through by the denominator to remove fractions. Then compare coefficients or substitute convenient values of $x$.

Example:

$$\frac{5x+1}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}$$

Multiply by $(x-2)(x+1)$:

$$5x+1=A(x+1)+B(x-2)$$

Expanding gives

$$5x+1=(A+B)x+(A-2B)$$

Now compare coefficients:

$$A+B=5$$

$$A-2B=1$$

Solving these gives

$$B=\frac{4}{3}$$

$$A=\frac{11}{3}$$

$$\frac{5x+1}{(x-2)(x+1)}=\frac{11/3}{x-2}+\frac{4/3}{x+1}$$

This is now easy to integrate.

Integrating the decomposed expression

Once the fraction has been split, use standard integration rules. The key rule is

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

So from the example above,

$$\int \frac{5x+1}{(x-2)(x+1)}\,dx=\int \frac{11/3}{x-2}\,dx+\int \frac{4/3}{x+1}\,dx$$

which gives

$$\frac{11}{3}\ln|x-2|+\frac{4}{3}\ln|x+1|+C$$

This result is typical of partial fraction integration: the answer often becomes a combination of logarithms.

Another common pattern appears with repeated factors. For example,

$$\int \frac{1}{x(x-1)^2}\,dx$$

can be decomposed as

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

After finding the constants, each term is integrated separately. The term $\frac{1}{(x-a)^2}$ integrates to $-\frac{1}{x-a}$, because

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

A full worked example

Let us integrate

$$\int \frac{2x+3}{x^2+x-2}\,dx$$

First factor the denominator:

$$x^2+x-2=(x+2)(x-1)$$

Now decompose:

$$\frac{2x+3}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}$$

Multiply by $(x+2)(x-1)$:

$$2x+3=A(x-1)+B(x+2)$$

Expand:

$$2x+3=(A+B)x+(-A+2B)$$

Compare coefficients:

$$A+B=2$$

$$-A+2B=3$$

From $A=2-B$, substitute into the second equation:

$$-(2-B)+2B=3$$

$$3B=5$$

$$B=\frac{5}{3}$$

Then

$$A=2-\frac{5}{3}=\frac{1}{3}$$

$$\frac{2x+3}{x^2+x-2}=\frac{1/3}{x+2}+\frac{5/3}{x-1}$$

Integrate:

$$\int \frac{2x+3}{x^2+x-2}\,dx=\frac{1}{3}\ln|x+2|+\frac{5}{3}\ln|x-1|+C$$

This example shows the full process: factor, decompose, solve for constants, then integrate.

Special case: irreducible quadratic factors

Sometimes the denominator contains a quadratic that cannot be factored further over the real numbers. For example,

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

The decomposition would be

$$\frac{x+1}{(x^2+1)(x-3)}=\frac{A}{x-3}+\frac{Bx+C}{x^2+1}$$

After solving for $A$, $B$, and $C$, the integral may produce both logarithmic and inverse tangent terms. This happens because

$$\int \frac{1}{x^2+1}\,dx=\arctan x+C$$

and more generally, linear numerators over quadratic denominators can often be split into parts that match a logarithm and an arctangent after rearranging.

For example, if you get

$$\int \frac{2x}{x^2+1}\,dx$$

then let

$$u=x^2+1$$

so that

$$du=2x\,dx$$

and the integral becomes

$$\int \frac{1}{u}\,du=\ln|u|+C=\ln(x^2+1)+C$$

This shows how partial fractions and substitution can both appear in the same topic.

Common mistakes to avoid

students, these are frequent errors to watch for:

forgetting to do polynomial division when the fraction is improper,
missing a term for a repeated factor,
using a constant numerator for a quadratic factor when it should be linear,
solving for constants incorrectly after multiplying by the denominator,
forgetting absolute values in logarithms, such as $\ln|x-a|$,
stopping too early before simplifying the algebra fully.

A good exam strategy is to check your decomposition by recombining the fractions. If the right-hand side simplifies back to the original fraction, your setup is likely correct ✅.

Conclusion

Partial fractions are a powerful calculus technique for integrating rational functions by rewriting them as simpler sums. In IB Mathematics: Analysis and Approaches HL, this method is important because it combines algebraic skill with integration methods. The process is: make the fraction proper, factor the denominator, set up the correct decomposition, find the constants, and then integrate term by term.

This topic fits naturally into calculus because it gives another way to find antiderivatives and helps connect algebraic structure to integration. It also supports further work in differential equations and modelling, where rational expressions often appear. If you can recognise the right structure and apply the method carefully, partial fractions become a reliable tool in your calculus toolkit 📘.

Study Notes

A rational function is a ratio of polynomials, such as $\frac{P(x)}{Q(x)}$.
Partial fractions are used to rewrite a rational function as a sum of simpler fractions.
First check whether the fraction is proper; if not, use polynomial division.
Factor the denominator completely before setting up the decomposition.
Distinct linear factors use constants like $\frac{A}{x-a}$.
Repeated linear factors need all powers, such as $\frac{A}{x-a}+\frac{B}{(x-a)^2}$.
Irreducible quadratic factors use linear numerators, such as $\frac{Ax+B}{x^2+1}$.
Multiply through by the denominator to solve for constants.
Integrals of the form $\int \frac{1}{x-a}\,dx$ give logarithms: $\ln|x-a|+C$.
Integrals of the form $\int \frac{1}{(x-a)^2}\,dx$ give $-\frac{1}{x-a}+C$.
Quadratic denominators may lead to arctangent terms, such as $\int \frac{1}{x^2+1}\,dx=\arctan x+C$.
Partial fractions are useful in calculus, differential equations, and modelling situations where rational expressions appear.