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Well, not quite. Long division turns out to be quite useful for evaluating integrals with quotients of polynomials. In this article, you will go over the definition of, formulas for, method for, and examples of integration using long division. It will not be that bad; we promise!
Definition of Integration of a Function Using the Long Division
So what is integration using long division anyway?
Integration using long division is a technique of integration that applies to integrals of the form
$$\int \frac{p(x)}{q(x)} \; dx, $$
where \(p(x)\) and \(q(x)\) are both polynomials and \(\deg(p(x))\geq\deg(q(x)).\)
It involves expressing \(p(x)\) as \(p(x) = q(x)s(x) + r(x)\) for polynomials \(s(x)\) and \(r(x)\) to simplify the integral:
$$\int \frac{p(x)}{q(x)} \; dx = \int s(x) \; dx + \int \frac{r(x)}{q(x)} \; dx.$$
The degree of a polynomial is its largest power. As examples, \(\deg(x^2 + 3x - 2) = 2\) and \(\deg(x^7 + 3) = 7\).
You could use integration using long division to evaluate
$$\int \frac{x^3 - 3x^2 + 2}{x - 3} \; dx$$
since \(\deg(x^3 - 3x^2 + 2) = 3\) and \(\deg(x-3) = 1\).
You would not use long division to evaluate
$$\int \frac{x^3 - 3x^2 + 2}{x^4 - 1} \; dx$$
since \(\deg(x^3 - 3x^2 + 2) = 3\) and \(\deg(x^4-1) = 4\).
Instead, you would probably use Integration by Parts. In this case, since the degree of the denominator is greater than the degree of the numerator, trying to use long division to get a polynomial \(s(x)\) would be like expecting \(\frac{5}{1000}\) to be an integer; the denominator is just too big for this to work.
Rules for Integration of a Function Using Long Division
There are a variety of techniques for integrating expressions that involve polynomials, including Trigonometric Substitution, long division, Integration by Partial Fractions, Integration by Substitution, Weierstrass substitution, and the power rule for integration (see Techniques of Integration). Let's go over rules and guidelines for when and how to use integration by long division, instead of these other techniques.
Guidelines for When to Use Integration by Long Division
The following table summarizes guidelines for determining when each of the integrations techniques discussed above is most appropriate.
| Integration Technique | When to Use | Example |
| Trigonometric Substitution | Integrand contains terms of the form \(\sqrt{x^2 + a^2}\), \(\sqrt{x^2 - a^2}\), \(\sqrt{a^2 - x^2}\), or (sometimes) the same terms without the square roots | $$\int \frac{\sqrt{x^2 - 4}}{x} dx$$ |
| Integration by Long Division | Integral is of the form$$\int \frac{p(x)}{q(x)} dx, $$where \(p(x), \, q(x)\) are both polynomials, \(\deg(p(x))\geq\deg(q(x))\), and no other technique applies | $$\int \frac{x^3-2}{x-3} dx$$ |
| Integration by Partial Fractions | Integral is of the form$$\int \frac{p(x)}{q(x)} dx, $$where \(p(x), \, q(x)\) are both polynomials, \(\deg(p(x))<\deg(q(x))\), and no other technique applies | $$\int \frac{x}{(x-2)(x-3)^2} dx$$ |
| Integration by Substitution | Integral is of the form$$\int f'(x) (a_nf(x)^n + ... a_0) dx;$$use the substitution \(u = f(x)\). | $$\int e^{x}(e^{3x} + 2) dx$$ |
| Power Rule for Integration (see Techniques of Integration) | Integrand is a simple polynomial | $$\int 3x^{-3} + 4x^{3/2} - 3 dx$$ |
| Weierstrass Substitution (see Techniques of Integration) | Integrand is a rational expression of trigonometric functions and u-substitution cannot be used | $$\int \frac{1}{\sin(x)} dx$$ |
These guidelines are not absolute, but they are a good place to start if you are not sure which technique of integration to use. In general, it is a good policy to avoid using trigonometric substitution, integration by long division, and integration by partial fractions unless absolutely necessary, as these techniques are more complicated than the others and easier to make mistakes with.
You may need to use several of these techniques for a single integral. For example, you can solve
$$\int e^{x}(e^{3x} + 2) dx$$
by first making the substitution \(u = e^{x}\), then using the power rule for integration. As always, it is important to keep an open mind when deciding which techniques of integration to use. There are often many ways to solve an integral, and the most elegant solution is not always the most obvious.
Rules for How to Use Integration by Long Division
There are a few rules you should follow when integrating a function using long division.
First, integration using long division only applies in cases where the integrand is a quotient of polynomials. It cannot be used otherwise.
Second, only integrate using long division when the degree of the numerator is greater than or equal to the degree of the denominator. If the degree of the numerator is less than the degree of the denominator, you must use another technique, like Integration by Partial Fractions, instead.
Formula for Integration of a Function Using Long Division
The formula that you use when using integration by long division is
$$\begin{align} \int \frac{p(x)}{q(x)} dx &= \int \frac{q(x)s(x) + r(x)}{q(x)} dx \\ \\ &= \int s(x) + \frac{r(x)}{q(x)} dx, \end{align}$$
where \(p(x) = s(x)q(x) + r(x).\) You find \(s(x)\) and \(r(x)\) using polynomial long division.
Polynomial Long Division
To divide a polynomial \(p(x)\) by a polynomial \(q(x)\), you must find the polynomials \(s(x), \, r(x)\) satisfying the formula
$$p(x) = q(x)s(x) + r(x).$$
You call \(r(x)\) the remainder term. This remainder term is just like the remainders you learned about when first studying fractions. For example, the remainder of \(\frac{10}{3}\) is \(1\), since
$$10 = 3(3) + 1.$$
Notice that the remainder is always less than the divisor; in this case, \(1 < 3.\) If \(p(x)\) is divisible by \(q(x)\), then \(r(x) = 0.\)
Polynomial long division is just like integer division. When you learned to divide integers, you probably did something like this:
Divide 273 by 15.
Solution: You can write the problem like this:
\begin{array}{r}\phantom{18)} \\15{\overline{\smash{\big)}\,273\phantom{)}}}\end{array}
First, multiply 15 by 10, putting a 1 in the tens place over the line. You can't multiply \(15\) by \(20\), since
$$15\cdot 20 = 300 > 273.$$
Since \(15\cdot 10 = 150\), you subtract 150 from 273, leaving 123.
\begin{array}{r}1\phantom{8)} \\15{\overline{\smash{\big)}\,273\phantom{)}}}\\\underline{-150\phantom{)}}\\123\phantom{)}\end{array}
Next, you multiply 15 by 8 and put an 8 in the units place over the line. Then, you subtract \(15\cdot 8 = 120\) from 123 to get 3. You can check that
$$15\cdot 9 = 135 > 123,$$
which is why you did not multiply 15 by anything greater than 8.
\begin{array}{r}18\phantom{)} \\15{\overline{\smash{\big)}\,273\phantom{)}}}\\\underline{-150\phantom{)}}\\123\phantom{)}\\\underline{-120\phantom{)}}\\3\phantom{)}\end{array}
15 does not divide 3, so you are done. So, \(273 = 15(18) + 3\). The remainder term is \(3\).
You could think of long division slightly differently, keeping track of place value.
Divide \(273\) by \(15\), this time keeping track of place value.
Solution: First, write out every number to keep track of its place value.
$$\begin{align}15 &= 1\cdot 10^1 + 5\cdot 10^0\\ \\273 &= 2\cdot 10^2 + 7\cdot 10^1 + 3\cdot 10^0\end{align}$$
This time, divide by 'getting rid of' one place value at a time, starting with the highest place value. The first term you want to 'get rid of' is the \(2\cdot 10^2\) term, since this is the largest term. To accomplish this, multiply \(1\cdot 10^1 + 5\cdot 10^0\) by \(2\cdot 10^1\) and subtract:
\begin{array}{r}2\cdot 10^1 \phantom{-3\cdot 10^0)} \\1\cdot 10^1+5\cdot 10^0{\overline{\smash{\big)}\,2\cdot 10^2 + 7\cdot 10^1 + 3\cdot 10^0\phantom{)}}}\\\underline{-~\phantom{(}(2\cdot 10^2 + 10\cdot 10^1)\phantom{7\cdot 10^0)}}\\-3\cdot 10^1 + 3\cdot 10^0\phantom{)}\end{array}
Now that all the \(10^2\) terms are gone, try to get rid of all the \(10^1\) terms. You can do this by multiplying \(1\cdot 10^1 + 5\cdot 10^0\) by \(-3\cdot 10^0\) and subtracting:
\begin{array}{r}2\cdot 10^1 -3\cdot 10^0\phantom{)} \\1\cdot 10^1+5\cdot 10^0{\overline{\smash{\big)}\,2\cdot 10^2 + 7\cdot 10^1 + 3\cdot 10^0\phantom{)}}}\\\underline{-~\phantom{(}(2\cdot 10^2 + 10\cdot 10^1)\phantom{3\cdot 10^0)}}\\-3\cdot 10^1 + 3\cdot 10^0\phantom{)}\\\underline{-~\phantom{(}(-3\cdot 10^1 - 15\cdot 10^0)}\\18\cdot 10^0\phantom{)}\end{array}
At this point, you are done, since you can't divide a \(10^0\) term by something containing a \(10^1\) term. The final answer is that
$$2\cdot 10^2 + 7\cdot 10^1 + 3\cdot 10^0 = (1\cdot 10^1 + 5\cdot 10^0)(2\cdot 10^1 - 3\cdot 10^0) + \frac{18\cdot 10^0}{1\cdot 10^1 + 5\cdot 10^0}.$$
Rewriting, this becomes:
$$273 = 15(20 - 3) + \frac{18}{15} = 15(17) + \frac{18}{15}.$$
You can check that this is equivalent to the answer you got before; the remainder term is just larger. This algorithm is a completely valid way to do integer division, but is not necessarily the best way to do it, since it sometimes gives you larger remainder terms than necessary.
However, this method would easily generalize to numbers written, say, in the form
\[a\cdot 2^3 + b\cdot 2^2 +c\cdot 2^1 + d\cdot 2^0\]
or
\[g\cdot 3^1 + h\cdot 3^0.\]
While the normal method of integer division is built to minimize the remainder, this method is built to keep track of place value, even if you aren't writing your numbers in base 10.
Polynomial long division is just like integer long division where you keep track of place value.
Divide \(2x^2 + 7x + 3\) by \(x + 5\).
Solution: You can write the problem like this:
\begin{array}{r}\phantom{2x-3)} \\x+5{\overline{\smash{\big)}\,2x^2 + 7x + 3\phantom{)}}}\end{array}
The term under the line is called the dividend, and the term to the left the divisor.
First, you want to get rid of the \(2x^2\) term. Forgetting about all the terms besides the \(2x^2\) term in the dividend and the \(x\) term in the divisor for a moment, we can notice that \(x(2x) = 2x^2\). So, multiply \(x+5\) by \(2x\), put \(2x\) in the \(x\) place above the line, and subtract \((x+5)(2x) = 2x^2 + 10x\) from \(2x^2 + 7x + 3.\)
\begin{array}{r}2x\phantom{-3)} \\x+5{\overline{\smash{\big)}\,2x^2 + 7x + 3\phantom{)}}}\\\underline{-~\phantom{(}(2x^2 + 10x)\phantom{-b)}}\\-3x+3\phantom{)}\end{array}
Now that you have gotten rid of the \(2x^2\) term, you want to get rid of the \(-3x\) term. So, multiply \(x+5\) by \(-3\), put \(-3\) in the units place above the line, and subtract \((x+5)(-3) = -3x - 15\) from \(-3x + 3.\)
\begin{array}{r}2x-3\phantom{)} \\x+5{\overline{\smash{\big)}\,2x^2 + 7x + 3\phantom{)}}}\\\underline{-~\phantom{(}(2x^2 + 10x)\phantom{-b)}}\\-3x+3\phantom{)}\\\underline{-~\phantom{()(}(-3x - 15)}\\\phantom{()}18\phantom{)}\end{array}
Since \(18\) has no \(x\) terms, \(x + 5\) cannot divide 18, and you are done. In this case, \(s(x) = 2x -3\) and your remainder term is \(r(x) = 18\). You can check that
$$2x^2 + 7x + 3 = (x+5)(2x-3) + 18.$$
Integration of a Function Using Long Division Method
To integrate
\[\int \frac{p(x)}{q(x)} dx\]
using long division, use the following method.
Determine what \(p(x)\) and \(q(x)\) are.
Using long division, divide \(p(x)\) by \(q(x)\) to get \(p(x) = q(x)s(x) + r(x)\) and rewrite the integral.
In symbols, \[\int \frac{p(x)}{q(x)} dx = \int s(x) + \frac{r(x)}{q(x)} dx.\]
Occasionally, it can be easier to factor \(p(x)\) and \(q(x)\) to look for common factors before resorting to long division.
Integrate.
If the integral is indefinite, don't forget to add \(C\)!
If the integral is definite, evaluate the antiderivative at the endpoints.
To integrate the resulting equation, you often must use Integration by Substitution, Integration by Partial Fractions, or both.
Examples of Integration of a Function Using Long Division
Let's do a few examples of integration using long division.
Let's evaluate \(\int \frac{x^3 - 3}{x - 2} dx\) using long division.
Step 1: The first step is to write down what \(p(x)\) and \(q(x)\) are. In this case, \(p(x) = x^3 - 3\) and \(q(x) = x-2\).
Step 2: Next, divide \(p(x)\) by \(q(x)\). To make division easier, write \(p(x)\) as \(p(x) = x^3 + 0x^2 + 0x - 3.\)
Don't forget this step; this is a common source of error!
\begin{array}{r}x^2+2x+4\phantom{)} \\x-2{\overline{\smash{\big)}\,x^3 + 0x^2 + 0x - 3\phantom{)}}}\\\underline{-(x^3-2x^2)\phantom{-bbbbbbb)}}\\2x^2+0x-3\phantom{)}\\\underline{-(2x^2-4x)\phantom{-b)}}\\4x-3\phantom{)}\\\underline{-(4x-8)}\\5\phantom{)}\end{array}
Thus, \(x^3 - 3 = (x-2)(x^2 + 2x + 4) + 5,\) and \(s(x) = x^2 + 2x + 4, \, r(x) = 5.\)
Next, rewrite the integral:
$$\begin{align}\int \frac{x^3 - 3}{x - 2} dx &= \int \frac{(x-2)(x^2 + 2x + 4) + 5}{x-2} dx\\ \\&= \int x^2 + 2x + 4 + \frac{5}{x-2} dx.\end{align}$$
Step 3: Finally, integrate:
$$\begin{align}\int \frac{x^3 - 3}{x - 2} dx &= \int x^2 + 2x + 4 + \frac{5}{x-2} dx\\ \\&= \int x^2 + 2x + 4 dx + \int \frac{5}{x-2} dx\\ \\&= \frac{1}{3}x^3 + x^2 + 4x + \int \frac{5}{x-2} dx\\ \\&= \frac{1}{3}x^3 + x^2 + 4x + 5\ln|x-2|.\end{align}$$
You can check either by differentiating or by using Integration by Substitution that
\[\int \frac{5}{x-2} dx = 5\ln|x-2|.\]
Let's do a slightly more complicated example.
Evaluate
\[\int \frac{4x^2 + 3x + 2}{x^2 + 5x - 3} dx.\]
Solution:
Step 1: First, identify \(p(x)\) and \(q(x)\). In this case, \(p(x) = 4x^2 + 3x + 2\) and \(q(x) = x^2 + 5x - 3\). Both \(p(x)\) and \(q(x)\) have degree 2, which is completely fine; you are still allowed to use long division.
Step 2: Next, divide \(p(x)\) by \(q(x)\).
\begin{array}{r}4\phantom{)} \\x^2 + 5x -3{\overline{\smash{\big)}\,4x^2 + 3x + 2\phantom{)}}}\\\underline{-(4x^2+20x - 12)\phantom{-b)}}\\-17x + 14\phantom{)}\end{array}
So, \(s(x) = 4\) and \(r(x) = -17x + 14\).
Step 3: Next, rewrite and evaluate your integral.
$$\begin{align}\int \frac{4x^2 + 3x + 2}{x^2 + 5x - 3} \; dx &= \int \frac{4(x^2 + 5x - 3) + (-17x + 14)}{x^2 + 5x - 3} \; dx\\ \\&= \int \frac{4}{x^2 + 5x - 3} \; dx + \int \frac{-17x + 14}{x^2 + 5x - 3} \; dx\\ \\&= \int \frac{4}{x^2 + 5x - 3} \; dx + \int \frac{-17x - 85/2 + 117/2}{x^2 + 5x - 3} \; dx\\ \\&= \int \frac{4}{x^2 + 5x - 3} \; dx - \frac{17}{2}\int \frac{2x+5}{x^2 + 5x - 3} \; dx \\ \\ & \quad + \frac{1}{2}\int\frac{117}{x^2 + 5x - 3} \; dx\\ \\&= 121\int \frac{1}{x^2 + 5x - 3} \; dx - \frac{17}{2}\int \frac{2x+5}{x^2 + 5x - 3} \; dx\end{align}$$
To evaluate the integral with \(2x + 5\) in the numerator, you can use Integration by Substitution with \(u = x^2 + 5x - 3\). To integrate the other term, you can use Integration by Partial Fractions:
$$ \begin{align} \int \frac{4x^2 + 3x + 2}{x^2 + 5x - 3} \; dx &= 121\int \frac{1}{x^2 + 5x - 3} \; dx - \frac{17}{2}\int \frac{2x+5}{x^2 + 5x - 3} \; dx\\ \\ &= 121\left(\frac{\ln(-2x + \sqrt{37} -5) - \ln(2x + \sqrt{37} + 5)}{\sqrt{37}}\right) \\ \\ & \quad -\frac{17}{2} \ln(x^2 + 5x - 3)\\ \\ &= \frac{121}{\sqrt{37}}\ln\left( \frac{ -2x + \sqrt{37} -5 }{2x + \sqrt{37} + 5 } \right) -\frac{17}{2} \ln(x^2 + 5x - 3) . \end{align} $$
Let's do one more example.
Evaluate \(\int_0^1 \frac{x^2 - 1}{x+1} \; dx\).
Solution: First, identify \(p(x)\) and \(q(x)\). In this case, \(p(x) = x^2 - 1\) is of degree 2 and \(q(x) = x+1\) is of degree 1, so you can use integration by long division.
Next, divide \(p(x)\) by \(q(x)\). To guard against division mistakes, rewrite \(p(x)\) as \(p(x) = x^2 + 0x - 1\).
\begin{array}{r}x-1\phantom{)} \\x+1{\overline{\smash{\big)}\,x^2 + 0x - 1\phantom{)}}}\\\underline{-(x^2+x)\phantom{-b)}}\\-x - 1\phantom{)}\\\underline{-\phantom{x^2 d}(-x-1)}\\0\phantom{)}\end{array}
So, \(s(x) = x-1\) and \(r(x) = 0\).
Next, rewrite your integral and evaluate:
$$\begin{align}\int_0^1 \frac{x^2 - 1}{x + 1} \; dx &= \int_0^1 \frac{(x-1)(x+1) + 0}{x+1} \; dx\\ \\&= \int_0^1 x - 1 \; dx\\ \\&= \frac{1}{2}x^2 - x \bigg|_0^1\\ \\&= \left(\frac{1}{2} - 1\right) - (0 - 0)\\ \\&= -\frac{1}{2}.\end{align}$$
Notice that, instead of using long division, you could have simply factored \(x^2 - 1\) as \((x-1)(x+1)\), divided, and integrated. In general, while long division (done properly) will always work for these kinds of integrals, checking if the numerator can be factored will often save you time and effort.
Integrating Functions Using Long Division - Key takeaways
- Integration by long division is used for integrals of the form \(\int \frac{p(x)}{q(x)}\), where p and q are polynomials and \(\deg(p(x))\geq \deg(q(x))\).
- Integration by long division is frequently used with other integration techniques like integration by substitution and partial fractions.
- Integration by long division should be used only when no more convenient methods apply.
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Frequently Asked Questions about Integration Using Long Division
What is the method of long division while using integration over a function?
Is a method used to simplify a rational function in order to integrate it more easily.
When to use long division in integration?
When there is a polynomial division, where the degree of the numerator is larger than the degree of the denominator.
What is an example of using long division in integration function?
An example coul be the integral of the functions:
g(x)/h(x).
Here:
g(x)= 3x3+x2-x-7
h(h)=x-1
What are the steps in performing long division in integration function?
- First, verify the degree of the polynomial in the division.
- Second, use long division to rewrite the integrand (factoring the polynomials may help in this step).
- Third, integrate the new integrand following the integrations rules.
How to use long division in integration function?
Always analyze if you have a division where the numerator has a degree bigger or equal to the denominator, if this is satisfied, calculate the long division of the polynomials..
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