Decomposition of rational functions

A way to reverse-engineer rational functions to make them easier to integrate

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Decomposition of rational functions is an algebraic trick that will allow you to re-express a complicated rational integrand as a sum of smaller rational functions. It's pretty straightforward when you know it, and easy to recognize when it's needed, but you need to practice it to learn a few tricks.

The basic idea is that any rational function with a denominator that is a product can be see to

Example 1

Integrate   $\int \frac{dx}{x^2 - 3x - 4}$

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Solution: Now this integrand has a factorable denominator, and that's crucial for the method of rational decomposition to work. Here's the factored integrand:

$$\frac{1}{x^2 - 3x - 4} = \frac{1}{(x + 1)(x - 4)}$$

Now imagine that fraction on the left to be just the sum of two simpler fractions.

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Cancelling single binomials after distributing the $(x+1)(x-4)$ on the right gives us a "basic equation." Notice that x can be anything, and that we're trying to find $A$ and $B$. We do that by convenient choice of $x$'s, like this:

$$1=A(x-4)+B(x+1)$$

"Basic equation," good for all $x$

Good choices are the ones that allow us to solve for $A$ and $B$, so $x = 4$ and $x = -1$ will do the trick:

$$ \begin{align} \text{let } \; x &= 4, \text{ then } \; 5B = 1\rightarrow B = \frac{1}{5} \\[5pt] \text{let } \; x &= -1, \text{ then } -5A = 1\rightarrow A = \frac{-1}{5} \end{align}$$

Now that we have $A$ and $B$, we can rewrite our integrand in two separate but simpler rational functions.

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

Example 2

Integrate   $\int \frac{x^2 - 5x + 6}{x^3 + 2x^2 + x}$

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Solution: Now this one is a little more complicated. We'll use a more complicated decomposition, which will be explained below. The denominator factors like this (note that the denominator can be factored:

$$\frac{x^2 - 5x + 6}{x^3 + 2x^2 + x} = \frac{x^2 - 5x + 6}{x(x + 1)^2}$$

The decomposition of the integrand is a bit trickier. We have to realize that the denominator, $x(x + 1)^2$, actually has three potential factors: $x$, $(x + 1)$ and $(x + 1)^2$. Not all may be necessary (i.e. one of the two coefficients below, $B$ or $C$ could be zero), but we need to find out by including all of the terms frorm the start:

$$\frac{x^2 - 5x + 6}{x(x + 1)^2} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}$$

We need that extra term under $C$ because the binomial ($x + 1$) is squared. Now we multiply both sides by the full denominator of the integrand ...

to get the basic equation:

$$x^2 - 5x + 6 = A(x + 1)^2 + Bx(x + 1) + Cx$$

Now in such an equation, $x$ can take on any value at all; it's $A$, $B$ and $C$ that we're trying to determine. Choosing convenient values of $x$ is the key. Choosing $x = 0$ and $x = -1$ will allow us to determine $A$ and $C$ because each leads to zeroing out of two of the other terms on the right.

$$ \begin{align} let \; x &= 0, \text{ then } 2A = 6, \; \rightarrow \; \bf A = 3 \\[5pt] let \; x &= -1, \text{ then } -C = -12, \; \rightarrow \; \bf C = 12 \end{align}$$

Now with $A$ and $C$ in hand, we can set $x$ equal to any other value and solve for $B$, but $x = 1$ is very convenient.

$$ \begin{align} let \; x = 1, \text{ then } 4A + 2B - 2C &= 2 \\[5pt] 4(3) + 2B - 2(-12) &= 2 \\[5pt] B &= 7 \end{align}$$

Note that $x$ could actually be anything in the determination of $B$ above, but $x = 1$ was convenient. With $A$, $B$ and $C$ in hand, we can re-write the integrand in a form that's easily integrable:

$$\frac{x^2 - 5x + 6}{x(x + 1)^2} = \frac{3}{x} + \frac{7}{x + 1} - \frac{12}{(x + 1)^2}$$

Now we just need to integrate. The first two are easy log integrals; the second and last are done by u-substitution (let u = x + 1 in either case).

$$\int \frac{x^2 - 5x + 6}{x(x + 1)^2} = 3 \int \frac{dx}{x} + 7 \int \frac{dx}{x + 1} - 12 \int \frac{dx}{(x + 1)^2}$$

The result is

$$= 3 ln|x| + 7 ln|x + 1| + \frac{12}{|1 + x|} + C$$

Example 3

Integrate   $\int \frac{2x^2 - x + 3}{x^3 + 4x} \, dx$

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Solution: The first thing to do is to factor the denominator, if possible. In this case we can pull out a common factor of $x$ from both terms:

$$\frac{2x^2 - x + 3}{x^3 + 4x} = \frac{2x^2 - x + 3}{x(x^2 + 4)}$$

Now the denominator of this rational function contains a quadratic ($x^2+4$) that can't be reduced (factored).The decomposition of the integrand this time looks like this:

$$\frac{2x^2 - x + 3}{x^3 + 4x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}$$

There are two "degrees of freedom" in the denominator on the right, $x^2+4$, where $x$ can take on two ± values and give the same result. The numerator of $Bx+C$ provides the second and third variables needed to completely specify the original function. Now multiplying by the original denominator gives

from which we get the basic equation:

$$2x^2 - x + 3 = A(x^2 + 4) + Bx^2 + Cx$$

Now choosing $x = 0$ and $x = -1$ will allow us to determine $A$ and $C$ because each leads to zeroing out of two of the other terms on the right.

$$let \; x = 0, \text{ then } 3 = 4A, \; \rightarrow \; A = \frac{3}{4}$$

In this case, finding $B$, is a little more complicated, but still doable. We set $x = -1$ and $x = 1$ to find two separate equations to solve for both $B$ and $C$:

And

$$let \; x = 2, \text{ then } 9 = 6 + 5 + C, \; \rightarrow \; C = -2$$

With $A$, $B$ and $C$ in hand, we can re-write the integrand in a form that's easily integrable:

$$\frac{2x^2 - x + 3}{x(x^2 + 4)} = \frac{3}{4x} + \frac{\frac{5}{4} x - 2}{x^2 + 4}$$

The integral is

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

Now we just need to integrate. The second can be done by making the substitution $u = x^2 + 4$, and the last is the derivative of $\text{tan}^{-1}(x)$. The result is

$$= \frac{3}{4} ln|x| + \frac{5}{4} ln|x^2 + 4| + \text{tan}^{-1} \left( \frac{x}{2} \right) + C$$

Example 4 — degree of numerator > degree of denominator

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

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Solution: The key to an integrand like this is to recognize that the degree of the denominator is less than the degree of the numerator, so we can simplify it by long division:

$$ \require{enclose} \require{cancel} \begin{array}{rll} x^2+x-3\phantom{0} \\[-3pt] x-1 \enclose{longdiv}{\cancel{x^3}+2x \phantom{0000}}\kern-.2ex \\[-3pt] \underline{{-(\cancel{x^3}-x^2)} \phantom{000}} \\[-3pt] -(\cancel{x^2}+2x) \phantom{000} \\[-3pt] \underline{{-(\cancel{x^2}-\; x)} \phantom{000}} \\[-3pt] \cancel{3x} \phantom{0000} \\[-3pt] \underline{{-(\cancel{3x}-3)}} \\[-3pt] 3 \;\; \end{array}$$

We can now substitute a polynomial plus a much simpler rational function (the remainder) for the original integrand:

Practice problems

Solve these integrals using decomposition of the rational integrand.

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1. $$\int \frac{x + 2}{x^2 - x - 6}\, dx$$

   Solution

   $$\int \frac{x + 2}{x^2 - x - 6}\, dx = \int \frac{x + 2}{(x - 3)(x + 2)} \, dx$$

   This function has a hole at x = -2. We can remove it by canceling the (x + 2) binomials and integrating the result:

   $$\int \frac{dx}{x - 3} = ln|x - 3| + C$$

   This integral didn't require any decomposition of the rational function!

2. $$\int \frac{x + 1}{x^2 - x - 6} \, dx$$

   Solution

   $$\int \frac{x + 1}{x^2 - x - 6} \, dx$$

   $$\frac{x + 1}{x^2 - x - 6} = \frac{x + 1}{(x - 3)(x + 2)}$$

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

   $$ \begin{align} x + 1 &= A(x + 2) + B(x - 3) \\[5pt] let \; x = -2, \; then \; -1 &= -5B \; \longrightarrow \; B = \frac{1}{5} \\[5pt] let \; x = 3, \; then \; 4 &= 5A \; \longrightarrow \; A = \frac{4}{5} \end{align}$$

   $$ \begin{align} \int \frac{x + 1}{x^2 - x - 6} \, dx &= \frac{4}{5} \int \frac{dx}{x - 3} \, dx + \frac{1}{5} \int \frac{dx}{x + 2} \, dx \\[5pt] &= \bf \frac{4}{5} ln|x - 3| + \frac{1}{5} ln|x + 2| + C \end{align}$$

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

   Solution

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

   Write the decomposition equation. (x - 1)² is a repeated factor, so it must be represented twice, as shown

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

   Write the basic equation and solve. Do the easiest stuff first.

   $$ \begin{align} x + 3 &= A(x - 1)^2 + Bx(x - 1) + Cx \\[5pt] let \; x &= 1, \; then \; C = 4 \\[5pt] let \; x &= 0, \; then \; A = 3 \\[5pt] let \; x &= -1, \; then \; 2 = 4A + 2B - C \\[5pt] B = \frac{2 - 4A - C}{2} &= \frac{2 - 4(3) - 4}{2} = -7 \end{align}$$

   Now set up the integrals and solve.

   $$ \begin{align} \frac{x + 3}{x(x - 1)^2} \, dx &= 3 \int \frac{dx}{x} - 7\int \frac{dx}{x - 1} + 4\int \frac{dx}{(x - 1)^2} \\[5pt] &= \bf 3 ln|x| - 7ln|x - 1| - \frac{4}{x - 1} + C \end{align}$$

   The last integral is done by u-substitution, u = x - 1, ...

4. $$\int \frac{dx}{x^2 + 6x + 8} \, dx$$

   Solution

   $$\int \frac{dx}{x^2 + 6x + 8} \, dx$$

   The denominator can be factored, always a good first step:

   $$\frac{1}{x^2 + 6x + 8} = \frac{1}{(x - 2)(x - 4)} = \frac{A}{x - 2} + \frac{B}{x - 4}$$

   Now write the basic equation and solve by picking strategic values of x:

   $$ \begin{align} A(x - 2) + B(x - 4) &= 1 \\[5pt] let \; x = 4, \; then \; B &= \frac{1}{2} \\[5pt] let \; x = 2, \; then \; A &= \frac{-1}{2} \\[5pt] \end{align}$$

   Now re-write the integral and solve:

   $$ \begin{align} \int \frac{dx}{x^2 + 6x + 8} \, dx &= \frac{1}{2}\int \frac{dx}{x - 4} - \frac{1}{2} \int \frac{dx}{x - 2}\\[5pt] &= ln|x - 4| - ln|x - 2| \\[5pt] &= ln \left| \frac{x - 4}{x - 2} \right| + C \end{align}$$

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

   Solution

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

   The denominator of the integrand is nicely factored in linear factors, so we can rewrite the rational function like this:

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

   Now write the basic equation and solve by picking strategic values of x:

   $$ \begin{align} A(2x + 1) + Bx &= 1 \\[5pt] let \; x = -\frac{1}{2}, \; then \; B &= -2 \\[5pt] let \; x = 0, \; then \; A &= 1 \\[5pt] \end{align}$$

   Now re-write the integral and solve:

   $$ \begin{align} \int \frac{dx}{x(2x + 1)} \, dx &= \int \frac{dx}{x} - 2\int \frac{dx}{2x + 1}\\[5pt] &= ln|x| - 2 \frac{1}{2}ln|2x + 1| \\[5pt] &= ln \left| \frac{x}{2x + 1} \right| + C \end{align}$$

   The second integral was done by u-substitution; that's where the factor of ½ came from.