6.9 Long division & completing the square

Achievable AP Calculus AB

6. Integration

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Long division & completing the square

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What you’ll learn

Algebraic manipulation: Rewrite complex integrals using long division, completing the square, or clever splitting to match standard forms.
Inverse trigonometric integrals: Evaluate advanced integrals that result in arcsine or arctangent functions.

Some integrals look complicated only because they are not in a standard form yet. These techniques rewrite them into familiar expressions so you can apply known rules. We focus on two tools:

Polynomial long division
Completing the square

Polynomial long division

Apply this technique when the rational function is improper, meaning:

$degree (numerator) \geq degree (denominator)$

This rewrites the fraction as a polynomial + a proper rational function.

Example 1.

$\int \frac{2 x ^{2} + 3 x + 5}{x + 1} d x$

Solution

Step 1: Divide

Using polynomial long division, we find:

$2 x^{2} \div x = 2 x$

Multiplying back gives $2 x (x + 1) = 2 x^{2} + 2 x$ , and subtracting leaves $x + 5$ .

Then,

$x \div x = 1$

Multiplying back gives $1 (x + 1) = x + 1$ , and subtracting leaves $4$ .

Thus,

$\frac{2 x ^{2} + 3 x + 5}{x + 1} = 2 x + 1 + \frac{4}{x + 1} .$

Step 2: Integrate term-by-term

$\int (2 x + 1 + \frac{4}{x + 1}) d x = x^{2} + x + 4 ln ∣ x + 1∣ + C$

Try the next example yourself before expanding the solution.

Example 2.

$\int \frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} d x$

Solution

(spoiler)

1. Polynomial long division:

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

2. Split the integrals:

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

3. Evaluate each term:

Term 1 (power rule):

$\int (x - 1) d x = \frac{1}{2} x^{2} - x$

Term 2 ( $u$ -sub):

Let $u = x^{2} + 1 ⟹ d u = 2 x d x$ .

$\int \frac{- x}{x ^{2} + 1} d x = - \int \frac{1}{u} \cdot \frac{1}{2} d u = - \frac{1}{2} ln (x^{2} + 1) + C$

Term 3 (inverse tangent):

Recall that $\frac{d}{d x} (arctan x) = \frac{1}{x ^{2} + 1}$ . Then:

$\int \frac{3}{x ^{2} + 1} d x = 3 arctan x + C$

Putting it all together,

$\int \frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} d x = \frac{1}{2} x^{2} - x - \frac{1}{2} ln (x^{2} + 1) + 3 arctan x + C$

Advanced inverse trig forms

Know the following integration formulas involving inverse trig functions, where:

$a$ is a constant ( $a > 0$ )
$u$ is a differentiable function of $x$

Integrals leading to inverse trig:

$\int \frac{1}{a ^{2} + u ^{2}} d u \int \frac{1}{a ^{2} - u ^{2}} d u = \frac{1}{a} arctan (\frac{u}{a}) + C = arcsin (\frac{u}{a}) + C$

Note: These formulas use $d u$ , not $d x$ . You must compute $d u$ from your substitution first.

Example 3.

$\int \frac{1}{4 + 9 x ^{2}} d x$

Solution

(spoiler)

Matching the terms,

Constant: $a^{2} = 4 ⟹ a = 2$
Function: $u^{2} = 9 x^{2} ⟹ u = 3 x$

With $u$ -substitution, $d u = 3 d x$ and

$\int \frac{1}{2 ^{2} + ( 3 x ) ^{2}} d x = \int \frac{1}{2 ^{2} + u ^{2}} \cdot \frac{1}{3} d u = \frac{1}{3} [\frac{1}{2} arctan (\frac{u}{2})] + C = \frac{1}{6} arctan (\frac{3 x}{2}) + C$

Completing the square

Apply this technique when the denominator contains a quadratic expression that cannot be factored. This rewrites the expression into a form that matches a standard inverse trigonometric integral.

Example 4.

$\int \frac{2}{x ^{2} + 4 x + 6} d x$

Solution

(spoiler)

1. Complete the square in the denominator:

$(x^{2} + 4 x) + 6 = (x^{2} + 4 x + 4) - 4 + 6 = (x + 2)^{2} + 2$

The integral rewritten:

$\int \frac{2}{( x + 2 ) ^{2} + 2} d x$

2. Recognize form:

$\int \frac{1}{a ^{2} + u ^{2}} d u = \frac{1}{a} arctan (\frac{u}{a}) + C$

We have:

Constant: $a^{2} = 2 ⟹ a = 2$
Function: $u = x + 2 ⟹ d u = d x$

Then:

$\int \frac{2}{( x + 2 ) ^{2} + 2} d x = 2 \int \frac{1}{u ^{2} + ( 2 ) ^{2}} d u = 2 [\frac{1}{2} arctan (\frac{u}{2})] + C = 2 arctan (\frac{x + 2}{2}) + C$

Example 5.

$\int \frac{1}{8 - 2 x - x ^{2}} d x$

Solution

(spoiler)

1. Complete the square in the denominator:

$8 - 2 x - x^{2} = 8 - (x^{2} + 2 x) = 8 - (x^{2} + 2 x + 1) + 1 = 9 - (x + 1)^{2}$

The integral rewritten:

$\int \frac{1}{9 - ( x + 1 ) ^{2}} d x$

2. Recognize form:

$\int \frac{1}{a ^{2} - u ^{2}} d u = arcsin (\frac{u}{a}) + C$

We have:

$a^{2} = 9 ⟹ a = 3$
$u = x + 1 ⟹ d u = d x$

Then:

$\int \frac{1}{9 - ( x + 1 ) ^{2}} d x = \int \frac{1}{3 ^{2} - u ^{2}} d u = arcsin (\frac{x + 1}{3}) + C$

Watch out for these common mistakes:

Always include $+ C$ on indefinite integrals. It’s easy to drop it in multi-step problems - make it a habit to write $+ C$ .
Convert the differential during $u$ -substitution. When you substitute $u$ , replace $d x$ with the correct $d u$ expression. Leaving $d x$ in a $u$ -integral is inconsistent and leads to errors in the coefficient.

Long division

Use when numerator degree ≥ denominator degree in rational functions
Rewrite as polynomial + proper fraction using polynomial long division
Integrate resulting terms separately:
- Polynomials: reverse power rule
- Proper fractions: $u$ -substitution or standard forms (e.g., logarithms, inverse trig)

Completing the square

Use for quadratics in denominators or under square roots that don’t factor easily
Rewrite quadratic as $(x + a)^{2} + b$ form
Allows matching to standard inverse trig integrals:
- $\int \frac{1}{x ^{2} + 1} d x = tan^{- 1} (x) + C$
- $\int \frac{1}{1 - x ^{2}} d x = sin^{- 1} (x) + C$
May require $u$ -substitution and adjusting constants for standard forms

Long division & completing the square

What you’ll learn

Algebraic manipulation: Rewrite complex integrals using long division, completing the square, or clever splitting to match standard forms.
Inverse trigonometric integrals: Evaluate advanced integrals that result in arcsine or arctangent functions.

Some integrals look complicated only because they are not in a standard form yet. These techniques rewrite them into familiar expressions so you can apply known rules. We focus on two tools:

Polynomial long division
Completing the square

Polynomial long division

Apply this technique when the rational function is improper, meaning:

$degree (numerator) \geq degree (denominator)$

This rewrites the fraction as a polynomial + a proper rational function.

Example 1.

$\int \frac{2 x ^{2} + 3 x + 5}{x + 1} d x$

Solution

Step 1: Divide

Using polynomial long division, we find:

$2 x^{2} \div x = 2 x$

Multiplying back gives $2 x (x + 1) = 2 x^{2} + 2 x$ , and subtracting leaves $x + 5$ .

Then,

$x \div x = 1$

Multiplying back gives $1 (x + 1) = x + 1$ , and subtracting leaves $4$ .

Thus,

$\frac{2 x ^{2} + 3 x + 5}{x + 1} = 2 x + 1 + \frac{4}{x + 1} .$

Step 2: Integrate term-by-term

$\int (2 x + 1 + \frac{4}{x + 1}) d x = x^{2} + x + 4 ln ∣ x + 1∣ + C$

Try the next example yourself before expanding the solution.

Example 2.

$\int \frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} d x$

Solution

(spoiler)

1. Polynomial long division:

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

2. Split the integrals:

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

3. Evaluate each term:

Term 1 (power rule):

$\int (x - 1) d x = \frac{1}{2} x^{2} - x$

Term 2 ( $u$ -sub):

Let $u = x^{2} + 1 ⟹ d u = 2 x d x$ .

$\int \frac{- x}{x ^{2} + 1} d x = - \int \frac{1}{u} \cdot \frac{1}{2} d u = - \frac{1}{2} ln (x^{2} + 1) + C$

Term 3 (inverse tangent):

Recall that $\frac{d}{d x} (arctan x) = \frac{1}{x ^{2} + 1}$ . Then:

$\int \frac{3}{x ^{2} + 1} d x = 3 arctan x + C$

Putting it all together,

$\int \frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} d x = \frac{1}{2} x^{2} - x - \frac{1}{2} ln (x^{2} + 1) + 3 arctan x + C$

Advanced inverse trig forms

Know the following integration formulas involving inverse trig functions, where:

$a$ is a constant ( $a > 0$ )
$u$ is a differentiable function of $x$

Integrals leading to inverse trig:

$\int \frac{1}{a ^{2} + u ^{2}} d u \int \frac{1}{a ^{2} - u ^{2}} d u = \frac{1}{a} arctan (\frac{u}{a}) + C = arcsin (\frac{u}{a}) + C$

Note: These formulas use $d u$ , not $d x$ . You must compute $d u$ from your substitution first.

Example 3.

$\int \frac{1}{4 + 9 x ^{2}} d x$

Solution

(spoiler)

Matching the terms,

Constant: $a^{2} = 4 ⟹ a = 2$
Function: $u^{2} = 9 x^{2} ⟹ u = 3 x$

With $u$ -substitution, $d u = 3 d x$ and

$\int \frac{1}{2 ^{2} + ( 3 x ) ^{2}} d x = \int \frac{1}{2 ^{2} + u ^{2}} \cdot \frac{1}{3} d u = \frac{1}{3} [\frac{1}{2} arctan (\frac{u}{2})] + C = \frac{1}{6} arctan (\frac{3 x}{2}) + C$

Completing the square

Apply this technique when the denominator contains a quadratic expression that cannot be factored. This rewrites the expression into a form that matches a standard inverse trigonometric integral.

Example 4.

$\int \frac{2}{x ^{2} + 4 x + 6} d x$

Solution

(spoiler)

1. Complete the square in the denominator:

$(x^{2} + 4 x) + 6 = (x^{2} + 4 x + 4) - 4 + 6 = (x + 2)^{2} + 2$

The integral rewritten:

$\int \frac{2}{( x + 2 ) ^{2} + 2} d x$

2. Recognize form:

$\int \frac{1}{a ^{2} + u ^{2}} d u = \frac{1}{a} arctan (\frac{u}{a}) + C$

We have:

Constant: $a^{2} = 2 ⟹ a = 2$
Function: $u = x + 2 ⟹ d u = d x$

Then:

$\int \frac{2}{( x + 2 ) ^{2} + 2} d x = 2 \int \frac{1}{u ^{2} + ( 2 ) ^{2}} d u = 2 [\frac{1}{2} arctan (\frac{u}{2})] + C = 2 arctan (\frac{x + 2}{2}) + C$

Example 5.

$\int \frac{1}{8 - 2 x - x ^{2}} d x$

Solution

(spoiler)

1. Complete the square in the denominator:

$8 - 2 x - x^{2} = 8 - (x^{2} + 2 x) = 8 - (x^{2} + 2 x + 1) + 1 = 9 - (x + 1)^{2}$

The integral rewritten:

$\int \frac{1}{9 - ( x + 1 ) ^{2}} d x$

2. Recognize form:

$\int \frac{1}{a ^{2} - u ^{2}} d u = arcsin (\frac{u}{a}) + C$

We have:

$a^{2} = 9 ⟹ a = 3$
$u = x + 1 ⟹ d u = d x$

Then:

$\int \frac{1}{9 - ( x + 1 ) ^{2}} d x = \int \frac{1}{3 ^{2} - u ^{2}} d u = arcsin (\frac{x + 1}{3}) + C$

Watch out for these common mistakes:

Always include $+ C$ on indefinite integrals. It’s easy to drop it in multi-step problems - make it a habit to write $+ C$ .
Convert the differential during $u$ -substitution. When you substitute $u$ , replace $d x$ with the correct $d u$ expression. Leaving $d x$ in a $u$ -integral is inconsistent and leads to errors in the coefficient.

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.