Integration Techniques:
Long Division & Completing the Square
Master two powerful algebraic tools that transform difficult rational integrands into forms you can integrate directly.
AP Calculus AB
Why Do We Need Algebraic Manipulation?
Not every integral fits a basic formula. When you encounter a rational function where the degree of the numerator is greater than or equal to the degree of the denominator, or a quadratic in the denominator that isn't a perfect square, standard rules fail. You need to rewrite the integrand first.
Two Essential Techniques
Polynomial Long Division
Use when \(\deg(\text{numerator}) \geq \deg(\text{denominator})\). Divide to get a polynomial plus a proper fraction.
Completing the Square
Use when the denominator is an irreducible quadratic. Rewrite to match \(\arctan\) or \(\arcsin\) forms.
Decision Flowchart
The Long Division Rule
After dividing, integrate term by term:
The Completing the Square Rule
This transforms denominators into forms matching these standard integrals:
Arctangent Form
\(\displaystyle\int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\!\frac{u}{a}+C\)
Arcsine Form
\(\displaystyle\int\frac{du}{\sqrt{a^2-u^2}}=\arcsin\!\frac{u}{a}+C\)
Long Division vs. Completing the Square
| Long Division | Completing the Square | |
|---|---|---|
| Trigger | \(\deg(\text{num}) \geq \deg(\text{den})\) | Irreducible quadratic in denominator |
| Goal | Reduce to proper fraction | Match \(\arctan\)/\(\arcsin\) form |
| Output | Polynomial + remainder fraction | \((u \pm h)^2 + k^2\) form |
| Common results | \(\ln|x|\), power rule | \(\arctan\), \(\arcsin\) |
Step-by-Step: Long Division
1
Compare degrees. If \(\deg(\text{numerator}) \geq \deg(\text{denominator})\), proceed.
2
Divide the leading term of the numerator by the leading term of the denominator.
3
Multiply the entire divisor by that result and subtract from the numerator.
4
Repeat until the remainder has degree less than the divisor.
5
Integrate each piece: the polynomial quotient term by term, plus the remainder fraction.
Step-by-Step: Completing the Square
1
Identify the quadratic \(ax^2 + bx + c\) in the denominator.
2
Factor out \(a\) from the first two terms: \(a(x^2 + \frac{b}{a}x) + c\).
3
Add and subtract \(\left(\frac{b}{2a}\right)^2\) inside the parentheses to form a perfect square.
4
Substitute \(u = x + \frac{b}{2a}\) and rewrite the integral.
5
Apply the \(\arctan\) or \(\arcsin\) formula to evaluate.
Example 1: Long Division
1
Divide: \(x^2 \div x = x\). Multiply: \(x(x+1) = x^2 + x\). Subtract: \((x^2+3)-(x^2+x) = -x+3\).
2
Continue: \(-x \div x = -1\). Multiply: \(-1(x+1) = -x-1\). Subtract: \((-x+3)-(-x-1) = 4\).
3
Result: \(\displaystyle\frac{x^2+3}{x+1} = x - 1 + \frac{4}{x+1}\)
Example 2: Cubic Over Linear
Performing long division of \(x^3 \div (x-2)\):
Example 3: Completing the Square
1
Complete the square: \(x^2 + 6x + 13 = (x+3)^2 + 4\)
2
Substitute: Let \(u = x + 3\), so \(du = dx\). The integral becomes \(\displaystyle\int \frac{du}{u^2 + 4}\).
3
Apply arctan formula with \(a = 2\):
Example 4: Numerator Adjustment
1
Complete the square: \(x^2 - 4x + 8 = (x-2)^2 + 4\)
2
Substitute: \(u = x - 2\), \(du = dx\). Integral becomes \(\displaystyle\int \frac{du}{u^2 + 4}\).
3
Result: \(\displaystyle\frac{1}{2}\arctan\!\frac{x-2}{2} + C\)
Example 5: Both Techniques Combined
1
Long division first (degrees are equal): \(\displaystyle\frac{x^2+1}{x^2+4x+8} = 1 + \frac{-4x - 7}{x^2+4x+8}\).
2
Complete the square on the denominator: \(x^2+4x+8 = (x+2)^2 + 4\).
3
Split the remainder: \(\displaystyle\frac{-4x-7}{(x+2)^2+4} = \frac{-4(x+2)+1}{(x+2)^2+4}\).
4
Integrate each piece: \(\displaystyle x - 2\ln\!\left[(x+2)^2+4\right] + \frac{1}{2}\arctan\!\frac{x+2}{2} + C\).
Common Mistakes
✕ Skipping Long Division
Attempting partial fractions or substitution on an improper rational function wastes time and leads to errors.
✕ Arithmetic Errors in CTS
Forgetting to subtract \(\left(\frac{b}{2a}\right)^2\) outside the squared term: you must both add and subtract it.
✕ Forgetting \(\frac{1}{a}\) in Arctan
Writing \(\arctan\frac{u}{a}\) without the \(\frac{1}{a}\) coefficient in front: the formula is \(\frac{1}{a}\arctan\frac{u}{a}\).
Summary of Integration Strategies
| Integrand Form | Technique | Result Pattern |
|---|---|---|
| \(\frac{x^3+\cdots}{x+a}\) | Long Division | Polynomial + \(\ln|x+a|\) |
| \(\frac{1}{x^2+bx+c}\) | Complete the Square | \(\frac{1}{a}\arctan\frac{u}{a}\) |
| \(\frac{1}{\sqrt{c-x^2-bx}}\) | Complete the Square | \(\arcsin\frac{u}{a}\) |
| \(\frac{x^2+\cdots}{x^2+bx+c}\) | Both: Divide then CTS | Mixed: polynomial + \(\ln\) + \(\arctan\) |
Question 1 of 5
What is the first step to evaluate \(\displaystyle\int \frac{x^2 + 5}{x - 1}\,dx\)?
Since \(\deg(x^2+5)=2 \geq \deg(x-1)=1\), you must perform long division first to reduce the integrand to a polynomial plus a proper fraction.
Question 2 of 5
Completing the square on \(x^2 + 8x + 20\) gives:
Half of 8 is 4, so \(x^2+8x+20 = (x+4)^2 - 16 + 20 = (x+4)^2+4\). Always compute \(c - (b/2)^2\) for the constant: \(20 - 16 = 4\).
Question 3 of 5
Evaluate \(\displaystyle\int \frac{x^2}{x + 3}\,dx\).
Long division: \(\frac{x^2}{x+3} = x - 3 + \frac{9}{x+3}\). Integrating: \(\frac{x^2}{2} - 3x + 9\ln|x+3| + C\). Option (D) is the division result, not the integral.
Question 4 of 5
What is \(\displaystyle\int \frac{dx}{x^2 + 2x + 5}\)?
Complete the square: \(x^2+2x+5 = (x+1)^2 + 4\). With \(u = x+1\) and \(a = 2\): \(\frac{1}{a}\arctan\frac{u}{a} = \frac{1}{2}\arctan\frac{x+1}{2} + C\). Don't forget the \(\frac{1}{a}\) factor.
Question 5 of 5
To evaluate \(\displaystyle\int \frac{x^2 + x + 1}{x^2 + 4}\,dx\), what should you do first?
Since \(\deg(\text{num}) = \deg(\text{den}) = 2\), perform long division first: \(\frac{x^2+x+1}{x^2+4} = 1 + \frac{x-3}{x^2+4}\). Then split the remainder and integrate each piece.