Introduction

This topic focuses on algebraic techniques that prepare integrals for evaluation. The essential knowledge tells us that techniques for finding antiderivatives include rearrangements into equivalent forms, such as long division and completing the square. These pre-integration steps transform complex integrands into forms we can integrate using basic formulas or substitution.

Long Division for Rational Functions

When to Use Long Division

Use long division when integrating a rational function $\frac{P ( x )}{Q ( x )}$ where the degree of the numerator is greater than or equal to the degree of the denominator.

Goal: Convert an improper rational function into a polynomial plus a proper rational function.

The Long Division Process

Example 1: Evaluate $\int \frac{x ^{3} + 2 x ^{2} + 1}{x ^{2} + 1}, d x$ .

Step 1: Perform long division since degree of numerator (3) ≥ degree of denominator (2).

$x^{2} + 1 x + 2 \overline{x^{3} + 2 x^{2} + 0 x + 1} \underline{x^{3} + 0 x^{2} + x} x^{3} + 2 x^{2} - x + 1 \underline{x^{3} + 2 x^{2} - 0 x + 2} x^{3} + 2 x^{2} - x - 1$

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

Step 2: Integrate term by term: $\int \frac{x ^{3} + 2 x ^{2} + 1}{x ^{2} + 1}, d x = \int (x + 2 + \frac{- x - 1}{x ^{2} + 1}), d x$

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

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

For $\int \frac{- x}{x ^{2} + 1}, d x$ : Let $u = x^{2} + 1$ , $d u = 2 x, d x$ $= - \frac{1}{2} ln ∣ x^{2} + 1∣$

For $\int \frac{- 1}{x ^{2} + 1}, d x$ : This is $- arctan (x)$

Final answer: $\frac{x ^{2}}{2} + 2 x - \frac{1}{2} ln (x^{2} + 1) - arctan (x) + C$

Simpler Long Division Examples

Example 2: Evaluate $\int \frac{x ^{2} + 3}{x + 1}, d x$ .

Perform long division:

$x + 1 x - 1 \overline{x^{2} + 0 x + 3} \underline{x^{2} + x} x^{2} + - x + 3 \underline{x^{2} + - x - 1} x^{2} + 0 x + 4$

Result: $\frac{x ^{2} + 3}{x + 1} = x - 1 + \frac{4}{x + 1}$

Integrate: $\int \frac{x ^{2} + 3}{x + 1}, d x = \int (x - 1 + \frac{4}{x + 1}), d x$

$= \frac{x ^{2}}{2} - x + 4 ln ∣ x + 1∣ + C$

Example 3: Evaluate $\int \frac{2 x ^{3} - x + 1}{x - 2}, d x$ .

Perform long division:

$x - 2 2 x^{2} + 4 x + 7 \overline{2 x^{3} + 0 x^{2} - x + 1} \underline{2 x^{3} - 4 x^{2}} 2 x^{3} + 4 x^{2} - x + 1 \underline{2 x^{3} + 4 x^{2} - 8 x} 2 x^{3} + 0 x^{2} + 7 x + 1 \underline{2 x^{3} + 0 x^{2} + 7 x - 14} 2 x^{3} + 0 x^{2} - x + 15$

Result: $\frac{2 x ^{3} - x + 1}{x - 2} = 2 x^{2} + 4 x + 7 + \frac{15}{x - 2}$

Integrate: $\int \frac{2 x ^{3} - x + 1}{x - 2}, d x = \int (2 x^{2} + 4 x + 7), d x + \int \frac{15}{x - 2}, d x$

$= \frac{2 x ^{3}}{3} + 2 x^{2} + 7 x + 15 ln ∣ x - 2∣ + C$

Completing the Square

When to Use Completing the Square

Use completing the square when the integrand involves:

A quadratic expression in the denominator
A quadratic under a square root
Forms that might lead to $arctan$ or $arcsin$

Goal: Rewrite $a x^{2} + b x + c$ in the form $a (x - h)^{2} + k$ to reveal patterns.

Basic Completing the Square Review

To complete the square for $x^{2} + b x + c$ :

Take half of the coefficient of $x$ : $\frac{b}{2}$
Square it: $(\frac{b}{2})^{2}$
Add and subtract this value

Example: $x^{2} + 6 x + 5 = (x^{2} + 6 x + 9) - 9 + 5 = (x + 3)^{2} - 4$

Inverse Tangent Integrals

Example 4: Evaluate $\int \frac{1}{x ^{2} + 4 x + 13}, d x$ .

Complete the square in the denominator: $x^{2} + 4 x + 13 = (x^{2} + 4 x + 4) + 9 = (x + 2)^{2} + 9$

Rewrite: $\int \frac{1}{( x + 2 ) ^{2} + 9}, d x$

Let $u = x + 2$ , $d u = d x$ : $= \int \frac{1}{u ^{2} + 9}, d u$

$= \int \frac{1}{9 ( \frac{u ^{2}}{9} + 1 )}, d u$

$= \frac{1}{9} \int \frac{1}{( \frac{u}{3} ) ^{2} + 1}, d u$

Let $v = \frac{u}{3}$ , $d v = \frac{1}{3} d u$ : $= \frac{1}{9} \cdot 3 \int \frac{1}{v ^{2} + 1}, d v$

$= \frac{1}{3} arctan (v) + C$

$= \frac{1}{3} arctan (\frac{u}{3}) + C$

$= \frac{1}{3} arctan (\frac{x + 2}{3}) + C$

Inverse Sine Integrals

Example 5: Evaluate $\int \frac{1}{- x ^{2} - 2 x + 8}, d x$ .

Complete the square under the square root: $- x^{2} - 2 x + 8 = - (x^{2} + 2 x) + 8$ $= - (x^{2} + 2 x + 1 - 1) + 8$ $= - (x^{2} + 2 x + 1) + 1 + 8$ $= - (x + 1)^{2} + 9$ $= 9 - (x + 1)^{2}$

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

Let $u = x + 1$ , $d u = d x$ : $= \int \frac{1}{9 - u ^{2}}, d u$

This is the arcsine form: $= arcsin (\frac{u}{3}) + C$

$= arcsin (\frac{x + 1}{3}) + C$

Logarithmic Forms After Completing the Square

Example 6: Evaluate $\int \frac{2 x + 1}{x ^{2} + 4 x + 5}, d x$ .

Complete the square in denominator: $x^{2} + 4 x + 5 = (x + 2)^{2} + 1$

Notice the numerator: $2 x + 1 = 2 (x + 2) - 3$

Split the integral: $\int \frac{2 x + 1}{x ^{2} + 4 x + 5}, d x = \int \frac{2 ( x + 2 ) - 3}{( x + 2 ) ^{2} + 1}, d x$

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

For the first integral, let $u = (x + 2)^{2} + 1$ : $\int \frac{2 ( x + 2 )}{( x + 2 ) ^{2} + 1}, d x = ln ∣ (x + 2)^{2} + 1∣ = ln (x^{2} + 4 x + 5)$

For the second integral: $\int \frac{3}{( x + 2 ) ^{2} + 1}, d x = 3 arctan (x + 2)$

Final answer: $ln (x^{2} + 4 x + 5) - 3 arctan (x + 2) + C$

Combining Both Techniques

Example 7: Evaluate $\int \frac{x ^{3}}{x ^{2} + 2 x + 5}, d x$ .

Step 1: Use long division (degree 3 ≥ degree 2):

$x^{2} + 2 x + 5 x - 2 \overline{x^{3} + 0 x^{2} + 0 x + 0} \underline{x^{3} + 2 x^{2} + 5 x} x^{3} + - 2 x^{2} - 5 x \underline{x^{3} + - 2 x^{2} - 4 x - 10} x^{3} + 0 x^{2} + - x + 10$

Result: $\frac{x ^{3}}{x ^{2} + 2 x + 5} = x - 2 + \frac{- x + 10}{x ^{2} + 2 x + 5}$

Step 2: Complete the square in the remaining fraction: $x^{2} + 2 x + 5 = (x + 1)^{2} + 4$

$\int \frac{x ^{3}}{x ^{2} + 2 x + 5}, d x = \int (x - 2), d x + \int \frac{- x + 10}{( x + 1 ) ^{2} + 4}, d x$

$= \frac{x ^{2}}{2} - 2 x + \int \frac{- ( x + 1 ) + 11}{( x + 1 ) ^{2} + 4}, d x$

By splitting the integral and using the inverse tangent formula, we get that the answer to this integral is $\frac{x ^{2}}{2} - 2 x - \frac{1}{2} ln ∣ x^{2} + 2 x + 5∣ + \frac{11}{2} arctan (\frac{x + 1}{2}) + C$ .

Common Patterns and Formulas

Key Integration Formulas After Rearrangement

After completing the square, common forms emerge:

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

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

Logarithm: $\int \frac{u ^{'}}{u}, d x = ln ∣ u ∣ + C$

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