6.10 - Integrating Functions Using Long Division and Completing the Square

Why Use These Techniques?

Sometimes the integrand can’t be integrated in the form it’s given to you. We use algebraic techniques to rewrite it into a form we recognize.

The integrand is everything between the

\int

and the

dx

\int \underbrace{f(x)}_{\text{integrand}} \, dx

Two main algebraic tools:

Polynomial long division (or synthetic division)

Completing the square

Both of these turn integrands into recognizable patterns.

Polynomial Long Division

When to Use It

Use long division when:

The numerator’s degree is greater than or equal to the denominator’s degree

You have a rational function (a polynomial divided by a polynomial)

Basic Procedure

Example:

\int \dfrac{3x^3 - x^2 - 5x + 1}{x - 2} \, dx

Step 1: Set up the long division.


          3x² + 5x + 5
       _______________
x - 2 | 3x³ - x² - 5x + 1

Step 2: Divide.

$x$ into $3x^3$ gives $3x^2$

Multiply: $3x^2(x-2) = 3x^3 - 6x^2$

Subtract: $(-x^2) - (-6x^2) = 5x^2$

Bring down: $5x^2 - 5x$

Keep going the same way.

Step 3: The result.

\frac{3x^3 - x^2 - 5x + 1}{x - 2} = 3x^2 + 5x + 5 + \frac{11}{x-2}

Step 4: Integrate.

\int \left(3x^2 + 5x + 5 + \frac{11}{x-2}\right) dx = x^3 + \frac{5x^2}{2} + 5x + 11\ln|x-2| + C

Note: for the

\frac{11}{x-2}

term, use

u = x-2

and

du = dx

to get

11\ln|x-2|

Synthetic Division

When You Can Use It

Synthetic division works ONLY when the divisor is linear in the form

(x - a)

, with no coefficient on the

x

Can use:

(x - 2)

(x + 5)

(x - 0)

Cannot use:

(2x - 1)

(x^2 + 1)

(3x + 7)

Procedure

Same example:

\dfrac{3x^3 - x^2 - 5x + 1}{x - 2}

Step 1: Identify the zero. For

x - 2

, the zero is

a = 2

Step 2: Set up the synthetic division.


2 | 3   -1   -5    1
  |      6   10   10
  |__________________
    3    5    5   11

How it works:

Bring down the 3

Multiply: $2 \times 3 = 6$ , add to $-1$ , get 5

Multiply: $2 \times 5 = 10$ , add to $-5$ , get 5

Multiply: $2 \times 5 = 10$ , add to $1$ , get 11 (the remainder)

Step 3: Interpret the results. The bottom row gives the coefficients, dropping one power each time:

3x^2 + 5x + 5 + \frac{11}{x-2}

Much faster than long division.

Example: Long Division with U-Substitution

Evaluate

\int \dfrac{6x^4 - 7x^3 + x^2 + 2x}{3x - 5} \, dx

Step 1: Long division.


           2x³ + x² + 2x + 4
        _________________________
3x - 5 | 6x⁴ - 7x³ + x² + 2x + 0

After dividing:

\frac{6x^4 - 7x^3 + x^2 + 2x}{3x - 5} = 2x^3 + x^2 + 2x + 4 + \frac{20}{3x-5}

Step 2: Integrate.

\int \left(2x^3 + x^2 + 2x + 4 + \frac{20}{3x-5}\right) dx

Step 3: Handle the fractional term with u-sub.

For

\int \dfrac{20}{3x-5} \, dx

u = 3x - 5, \quad du = 3 \, dx, \quad dx = \frac{du}{3}

\int \frac{20}{u} \cdot \frac{du}{3} = \frac{20}{3}\ln|u| = \frac{20}{3}\ln|3x-5|

Step 4: The complete answer.

= \frac{x^4}{2} + \frac{x^3}{3} + x^2 + 4x + \frac{20}{3}\ln|3x-5| + C

Completing the Square

The Formula

For

x^2 + bx

, complete the square by adding and subtracting

\left(\frac{b}{2}\right)^2

\boxed{x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2}

Steps:

Take half of the linear coefficient

Square it

Add it to create the perfect square

Subtract it to keep the expression balanced

Example 1: Basic Completing the Square

Evaluate

\int \dfrac{1}{x^2 + 6x + 10} \, dx

Step 1: Focus on the denominator:

x^2 + 6x + 10

Step 2: Complete the square.

Half of 6 is 3, and

3^2 = 9

, so

x^2 + 6x + 10 = (x^2 + 6x + 9) - 9 + 10 = (x+3)^2 + 1

Step 3: Rewrite the integral.

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

Step 4: Recognize the pattern. This matches

\int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u) + C

With

u = x + 3

and

du = dx

= \tan^{-1}(x+3) + C

Example 2: With a Negative Leading Coefficient

Evaluate

\int \dfrac{1}{\sqrt{-x^2 - 8x - 15}} \, dx

Step 1: Factor the negative out of the quadratic.

-x^2 - 8x - 15 = -(x^2 + 8x + 15)

Step 2: Complete the square.

Half of 8 is 4, and

4^2 = 16

, so

x^2 + 8x + 15 = (x^2 + 8x + 16) - 16 + 15 = (x+4)^2 - 1

Step 3: Substitute back.

-(x^2 + 8x + 15) = -[(x+4)^2 - 1] = 1 - (x+4)^2

Step 4: Rewrite the integral.

\int \frac{1}{\sqrt{1 - (x+4)^2}} \, dx

Step 5: Recognize the inverse trig pattern. This matches

\int \frac{1}{\sqrt{1-u^2}} \, du = \sin^{-1}(u) + C

With

u = x + 4

= \sin^{-1}(x+4) + C

Complex Example: Multiple Techniques

Evaluate

\int \dfrac{32}{x^2 - 4x + 20} \, dx

Step 1: Complete the square.

x^2 - 4x + 20 = (x^2 - 4x + 4) - 4 + 20 = (x-2)^2 + 16

Step 2: Rewrite.

\int \frac{32}{(x-2)^2 + 16} \, dx

Step 3: Factor to get a “+1.” We need the form

u^2 + 1

for

\tan^{-1}

, so factor out the 16:

= \int \frac{32}{16\left[\frac{(x-2)^2}{16} + 1\right]} \, dx = \int \frac{2}{\left(\frac{x-2}{4}\right)^2 + 1} \, dx

Step 4: U-substitution.

u = \frac{x-2}{4}, \quad du = \frac{1}{4} \, dx, \quad dx = 4 \, du

Step 5: Substitute.

\int \frac{2}{u^2 + 1} \cdot 4 \, du = 8\int \frac{1}{u^2 + 1} \, du

Step 6: Integrate.

= 8\tan^{-1}(u) + C = 8\tan^{-1}\left(\frac{x-2}{4}\right) + C

Definite version: the same antiderivative evaluates a definite integral. For example:

\int_2^6 \frac{32}{x^2 - 4x + 20} \, dx = 8\tan^{-1}\left(\frac{x-2}{4}\right) \Bigg|_2^6 = 8\left[\tan^{-1}(1) - \tan^{-1}(0)\right] = 8 \cdot \frac{\pi}{4} = 2\pi

Key Takeaways

Use long division when the numerator’s degree is greater than or equal to the denominator’s

Synthetic division is faster, but it only works for $(x - a)$ divisors

The remainder becomes $\frac{R}{\text{divisor}}$ in the result

Completing the square creates perfect square trinomials

Formula: $x^2 + bx$ becomes $\left(x + \frac{b}{2}\right)^2$ after adding and subtracting $\left(\frac{b}{2}\right)^2$

Factor out negatives before completing the square, since you need the $x^2$ coefficient to be 1

Goal: transform the integrand into a recognizable pattern, especially inverse trig

Inverse trig patterns: look for $u^2 + 1$ , $\sqrt{1-u^2}$ , and $u\sqrt{u^2-1}$

Combine techniques. You often need completing the square plus u-substitution

Check the degree and make sure you’re in proper form before integrating

These are some of the most challenging integration techniques in the course

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