Integrating with Long Division and Completing the Square

Some rational functions don’t fit any basic rule or an obvious substitution as they stand. Two algebra tools from earlier courses fix that. Long division splits a “top-heavy” fraction into a polynomial plus a simpler fraction. Completing the square turns a quadratic denominator into the form $(x - h)^2 + a^2$, which integrates to an arctangent.

Key ideas

Two integrals you’ll use constantly

For a linear denominator (from a quick substitution $u = ax + b$):

$$\int \frac{1}{ax + b}\,dx = \frac{1}{a}\ln|ax + b| + C$$

For a sum of squares (from $\int \tfrac{1}{1 + u^2}\,du = \arctan u + C$ with $u = \tfrac{x}{a}$):

$$\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C \qquad (a \gt 0)$$

Here’s where the second one comes from. Factor out $a^2$ and let $u = \tfrac{x}{a}$, so $dx = a\,du$:

$$\int \frac{1}{a^2\left(1 + \frac{x^2}{a^2}\right)}\,dx = \frac{1}{a^2}\int \frac{a\,du}{1 + u^2} = \frac{1}{a}\arctan u + C$$

Long division first

If the degree of the numerator is greater than or equal to the degree of the denominator, divide first. The result is

$$\frac{\text{numerator}}{\text{denominator}} = \text{quotient} + \frac{\text{remainder}}{\text{denominator}},$$

and both pieces are easy to integrate. You can use long division, synthetic division, or just rewrite the numerator cleverly (for example, $\frac{x + 3}{x + 1} = \frac{(x + 1) + 2}{x + 1} = 1 + \frac{2}{x + 1}$).

Completing the square

If the denominator is a quadratic with no real zeros (its discriminant is negative), complete the square:

$$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right)$$

Then substitute $u = x + \tfrac{b}{2}$ (so $du = dx$) and use the arctangent integral.

If the numerator is a linear term, first split off a multiple of the derivative of the denominator (which gives a logarithm), and complete the square on what’s left (see Practice 9).

Choosing a technique

What you see	What to do
Numerator degree ≥ denominator degree	Long division first
Numerator is a multiple of the denominator’s derivative	Substitution, giving a $\ln$
$1$ over a quadratic with no real zeros	Complete the square, giving an $\arctan$

In AP Calculus AB, only these cases come up. Denominators that factor into distinct (non-repeating) linear factors usually need partial fractions, which is BC-only in AP.

Worked examples

Example 1: Long division

Find $\displaystyle\int \frac{x^2 + 3x + 1}{x + 1}\,dx$.

Solution. The top has degree $2$ and the bottom degree $1$, so divide. Since $(x + 1)(x + 2) = x^2 + 3x + 2$,

$$x^2 + 3x + 1 = (x + 1)(x + 2) - 1 \quad\Rightarrow\quad \frac{x^2 + 3x + 1}{x + 1} = x + 2 - \frac{1}{x + 1}.$$ $$\int \left(x + 2 - \frac{1}{x + 1}\right)dx = \frac{x^2}{2} + 2x - \ln|x + 1| + C$$

Example 2: The arctangent form

Find (a) $\displaystyle\int \frac{1}{x^2 + 9}\,dx$ and (b) $\displaystyle\int \frac{1}{4x^2 + 1}\,dx$.

Solution.

(a) This is $\dfrac{1}{x^2 + a^2}$ with $a = 3$:

$$\int \frac{1}{x^2 + 9}\,dx = \frac{1}{3}\arctan\!\left(\frac{x}{3}\right) + C$$

(b) Write $4x^2 = (2x)^2$ and let $u = 2x$, so $dx = \tfrac{1}{2}\,du$:

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

Example 3: Completing the square

Find $\displaystyle\int \frac{1}{x^2 + 6x + 13}\,dx$.

Solution. The discriminant is $36 - 52 \lt 0$, so the denominator doesn’t factor. Complete the square:

$$x^2 + 6x + 13 = (x^2 + 6x + 9) + 4 = (x + 3)^2 + 2^2$$

Let $u = x + 3$, $du = dx$:

$$\int \frac{1}{u^2 + 2^2}\,du = \frac{1}{2}\arctan\!\left(\frac{u}{2}\right) + C = \frac{1}{2}\arctan\!\left(\frac{x + 3}{2}\right) + C$$

Example 4: A definite integral

Evaluate $\displaystyle\int_1^3 \frac{1}{x^2 - 2x + 5}\,dx$.

Solution. Complete the square: $x^2 - 2x + 5 = (x - 1)^2 + 4$. With $u = x - 1$, the limits become $u = 0$ and $u = 2$:

$$\int_0^2 \frac{1}{u^2 + 4}\,du = \left[\frac{1}{2}\arctan\frac{u}{2}\right]_0^2 = \frac{1}{2}\arctan 1 - \frac{1}{2}\arctan 0 = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}$$

(Arctangent values are in radians: $\arctan 1 = \tfrac{\pi}{4}$.)

Common mistakes

Skipping the division. $\int \frac{x^2}{x + 1}\,dx$ is not $\frac{x^3}{3}\ln|x + 1|$ or anything like it. When the top’s degree is at least the bottom’s, divide first.

Forgetting the remainder. After dividing, the remainder term is usually where the logarithm comes from. Check your division by multiplying back.

Losing the 1/a. $\int \frac{1}{x^2 + 9}\,dx = \frac{1}{3}\arctan\frac{x}{3} + C$. Both the $\frac{1}{3}$ in front and the $\frac{x}{3}$ inside are needed.

Using ln for every fraction. $\int \frac{1}{x^2 + 4}\,dx$ is not $\ln(x^2 + 4)$; the numerator isn’t the derivative of the denominator. A logarithm appears only when the top is a multiple of the bottom’s derivative.

Errors completing the square. Half the $x$-coefficient, square it, add and subtract. For $x^2 - 2x + 5$: $(x - 1)^2 - 1 + 5 = (x - 1)^2 + 4$.

Practice

1. (Warm-up) Find $\displaystyle\int \frac{3}{x - 5}\,dx$.

Solution

$$3\ln|x - 5| + C$$

2. (Warm-up) Find $\displaystyle\int \frac{1}{x^2 + 16}\,dx$.

Solution

With $a = 4$:

$$\frac{1}{4}\arctan\!\left(\frac{x}{4}\right) + C$$

3. (Warm-up) Find $\displaystyle\int \frac{x + 3}{x + 1}\,dx$.

Solution

$\dfrac{x + 3}{x + 1} = \dfrac{(x + 1) + 2}{x + 1} = 1 + \dfrac{2}{x + 1}$:

$$\int \left(1 + \frac{2}{x + 1}\right)dx = x + 2\ln|x + 1| + C$$

4. (Core) Find $\displaystyle\int \frac{x^2 - 4x + 7}{x - 2}\,dx$.

Solution

$x^2 - 4x + 7 = (x - 2)^2 + 3 = (x - 2)(x - 2) + 3$, so

$$\frac{x^2 - 4x + 7}{x - 2} = x - 2 + \frac{3}{x - 2}.$$$$\int \left(x - 2 + \frac{3}{x - 2}\right)dx = \frac{x^2}{2} - 2x + 3\ln|x - 2| + C$$

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

Solution

The degrees are equal, so divide: $\dfrac{x^2 + 2}{x^2 + 1} = \dfrac{(x^2 + 1) + 1}{x^2 + 1} = 1 + \dfrac{1}{x^2 + 1}$.

$$\int \left(1 + \frac{1}{x^2 + 1}\right)dx = x + \arctan x + C$$

6. (Core) Find $\displaystyle\int \frac{1}{x^2 + 4x + 8}\,dx$.

Solution

$x^2 + 4x + 8 = (x + 2)^2 + 4$. With $u = x + 2$:

$$\int \frac{1}{u^2 + 2^2}\,du = \frac{1}{2}\arctan\!\left(\frac{x + 2}{2}\right) + C$$

7. (Core) Find $\displaystyle\int \frac{x^3 + 2x^2 + 1}{x + 2}\,dx$.

Solution

$x^3 + 2x^2 = x^2(x + 2)$, so $x^3 + 2x^2 + 1 = x^2(x + 2) + 1$ and

$$\frac{x^3 + 2x^2 + 1}{x + 2} = x^2 + \frac{1}{x + 2}.$$$$\int \left(x^2 + \frac{1}{x + 2}\right)dx = \frac{x^3}{3} + \ln|x + 2| + C$$

8. (Challenge) Evaluate $\displaystyle\int_0^1 \frac{x^3}{x^2 + 1}\,dx$.

Solution

Divide: $x^3 = x(x^2 + 1) - x$, so $\dfrac{x^3}{x^2 + 1} = x - \dfrac{x}{x^2 + 1}$.

For the second piece, $u = x^2 + 1$ gives $\int \dfrac{x}{x^2 + 1}\,dx = \tfrac{1}{2}\ln(x^2 + 1)$.

$$\int_0^1 \left(x - \frac{x}{x^2 + 1}\right)dx = \left[\frac{x^2}{2} - \frac{1}{2}\ln(x^2 + 1)\right]_0^1 = \frac{1}{2} - \frac{1}{2}\ln 2 \approx 0.153$$

9. (Challenge) Find $\displaystyle\int \frac{2x + 6}{x^2 + 2x + 5}\,dx$.

Solution

The derivative of the denominator is $2x + 2$. Split the numerator: $2x + 6 = (2x + 2) + 4$.

$$\int \frac{2x + 2}{x^2 + 2x + 5}\,dx + \int \frac{4}{x^2 + 2x + 5}\,dx$$

The first is a logarithm ($u = x^2 + 2x + 5$): $\ln(x^2 + 2x + 5)$. No absolute value is needed, because $x^2 + 2x + 5 = (x + 1)^2 + 4$ is always positive.

For the second, complete the square: $\displaystyle\int \frac{4}{(x + 1)^2 + 4}\,dx = 4 \cdot \frac{1}{2}\arctan\!\left(\frac{x + 1}{2}\right)$.

$$\int \frac{2x + 6}{x^2 + 2x + 5}\,dx = \ln(x^2 + 2x + 5) + 2\arctan\!\left(\frac{x + 1}{2}\right) + C$$