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Calculating Integrals of Fractions with Quadratic Denominators

Reviewed by Calculator Editorial Team

Integrals of fractions with quadratic denominators are common in calculus problems. This guide explains the methods for solving such integrals, including substitution, completing the square, and partial fraction decomposition.

Introduction

Integrals of the form ∫(A x + B)/(C x² + D x + E) dx are frequently encountered in calculus. These integrals can be approached using several methods depending on the nature of the quadratic denominator.

The general approach involves:

  1. Checking if the denominator can be factored into linear terms
  2. Using substitution if the numerator is a linear function of the denominator
  3. Completing the square if the denominator is a perfect square
  4. Using partial fractions if the denominator can be factored

Basic Methods

Substitution Method

When the numerator is a linear function of the denominator, substitution can simplify the integral. For example:

∫(2x + 3)/(x² + 4x + 5) dx

Let u = x² + 4x + 5, then du = (2x + 4) dx. Notice that the numerator is (2x + 3), which is similar to du but not identical. We can adjust:

∫(2x + 4)/(x² + 4x + 5) dx + ∫(-1)/(x² + 4x + 5) dx = ln|x² + 4x + 5| - ∫1/(x² + 4x + 5) dx

Completing the Square

When the denominator is a perfect square, completing the square can simplify the integral. For example:

∫1/(x² + 2x + 2) dx

Complete the square in the denominator:

x² + 2x + 2 = (x + 1)² + 1

Now the integral becomes:

∫1/((x + 1)² + 1) dx

This can be solved using the standard integral formula for arctangent.

Partial Fractions

When the denominator can be factored into linear terms, partial fraction decomposition is an effective method. For example:

∫(x + 2)/(x² - x - 6) dx

First factor the denominator:

x² - x - 6 = (x - 3)(x + 2)

Express the integrand as partial fractions:

(x + 2)/(x² - x - 6) = A/(x - 3) + B/(x + 2)

Solve for A and B, then integrate each term separately.

Worked Examples

Example 1: Substitution Method

Calculate ∫(3x + 2)/(x² + 4x + 5) dx

  1. Let u = x² + 4x + 5, du = (2x + 4) dx
  2. Adjust the numerator: (3x + 2) = (2x + 4) - 2
  3. Integrate: ∫(2x + 4)/(x² + 4x + 5) dx - 2∫1/(x² + 4x + 5) dx
  4. First integral: ln|x² + 4x + 5|
  5. Second integral: complete the square in denominator
  6. Final result: ln|x² + 4x + 5| - 2∫1/((x + 2)² + 1) dx

Example 2: Partial Fractions

Calculate ∫(2x + 1)/(x² - 4) dx

  1. Factor denominator: x² - 4 = (x - 2)(x + 2)
  2. Express as partial fractions: (2x + 1)/(x² - 4) = A/(x - 2) + B/(x + 2)
  3. Solve for A and B: A = 1, B = 1
  4. Integrate: ∫1/(x - 2) dx + ∫1/(x + 2) dx = ln|x - 2| + ln|x + 2| + C

FAQ

When should I use substitution instead of partial fractions?

Use substitution when the numerator is a linear function of the denominator. This method is simpler and more direct than partial fractions in such cases.

What if the denominator doesn't factor nicely?

If the denominator doesn't factor into linear terms, you may need to complete the square or use other techniques like integration by parts.

How do I know if the denominator is a perfect square?

Check if the discriminant (D² - 4CE) of the quadratic is zero. If it is, the denominator is a perfect square.