Question

Use the method of partial fractions to evaluate each of the following integrals. \(\int \frac{2 x^{2}+4 x+22}{x^{2}+2 x+10} d x\)

Short answer

The integral evaluates to: \(2x - \frac{1}{2}\ln|x^2 + 2x + 10| + \frac{1}{3}\arctan\left(\frac{x+1}{3}\right) + C\).

Step-by-step solution

Check for Long Division

Since the degree of the numerator \(2x^2 + 4x + 22\) matches the degree of the denominator \(x^2 + 2x + 10\), we need to perform long division. Divide \(2x^2 + 4x + 22\) by \(x^2 + 2x + 10\) to get: \(2\) as the quotient, and \(-x + 2\) as the remainder.

Re-write the Integral

Rewrite the integral using the result from long division: \(\int \frac{2x^2 + 4x + 22}{x^2 + 2x + 10} dx = \int 2dx + \int \frac{-x + 2}{x^2 + 2x + 10} dx\). Now solve each integral separately.

Integrate the First Term

The first integral, \(\int 2dx\), is straightforward: \(2x + C_1\), where \(C_1\) is a constant of integration.

Complete the Square

First, rewrite the denominator using completing the square: \(x^2 + 2x + 10 = (x + 1)^2 + 9\). This will help in simplifying the integration process for the second term.

Use Substitution for the Second Integral

Let \(u = x + 1\) which implies \(du = dx\). The integral becomes \(\int \frac{-u + 1}{u^2 + 9} du\). Break the fraction: \(-\int \frac{u}{u^2 + 9} du + \int \frac{1}{u^2 + 9} du\).

Integrate Using Logarithmic and Inverse Trigonometric Forms

The first part \(-\int \frac{u}{u^2 + 9} du\) integrates to \(-\frac{1}{2}\ln|u^2 + 9| + C_2\). The second part \(\int \frac{1}{u^2 + 9} du\) integrates to \(\frac{1}{3}\arctan\left(\frac{u}{3}\right) + C_3\).

Combine All Parts

Combine all the integrated parts with the constant terms: \(2x - \frac{1}{2}\ln|x^2 + 2x + 10| + \frac{1}{3}\arctan\left(\frac{x+1}{3}\right) + C\), where \(C\) is a constant of integration representing \(C_1, C_2, C_3\).

Key concepts

long division

Long division is a crucial step when dealing with partial fraction integration, especially when the degree of the numerator is equal to or greater than the degree of the denominator. In simplified terms, long division helps us break down complex fractions into simpler, more manageable parts. Here's how it works:

• If the degree of the numerator is greater than or equal to the denominator, perform long division.
• Divide each term of the numerator by the highest degree term in the denominator.
• Subtract the result from the original fraction to find the remainder.

With our example, dividing the numerator \(2x^2 + 4x + 22\) by the denominator \(x^2 + 2x + 10\), we find the quotient is \(2\) and the remainder is \(-x + 2\). This allows us to rewrite the integral into a more straightforward form, splitting it into two separate integrals.

completing the square

Completing the square is a technique used to make quadratic expressions easier to integrate, especially when dealing with inverse trigonometric functions. By transforming a quadratic polynomial, we alter it to a form that reveals a perfect square plus a constant, which is easier to work with.

• Identify the quadratic expression, such as \(x^2 + 2x + 10\).
• Create a perfect square trinomial. For this, focus on the \(x\) terms: \(x^2 + 2x\). Half the linear coefficient, \(1\), square it, and add inside the expression making it \((x + 1)^2\).
• Add the squared term to complete the expression: \((x + 1)^2 + 9\).

This newly completed square form simplifies the integral, preparing it for methods like substitution or trigonometric integration.

substitution

Substitution is useful when integrals seem complex at first glance, often simplifying them through the introduction of a new variable. Think of it as changing the perspective of the problem.

• Choose a substitution that makes the integral easier. Here, substitute \(u = x + 1\), simplifying the denominator as \(u^2 + 9\).
• Find the differential, \(du = dx\), using it to change all instances of \(x\) in terms of \(u\).
• Rewrite the integral fully in terms of \(u\): \(-\int \frac{u}{u^2 + 9} du + \int \frac{1}{u^2 + 9} du\).

Substitution helps redefine and simplify the setup, allowing for more straightforward integration.

trigonometric integration

Trigonometric integration is a strategy for integrating functions involving trigonometric terms. It utilizes known integrals of trigonometric expressions and identities.

• Use the inverse trigonometric identities: For example, \(\int \frac{1}{u^2 + a^2} du\) links to \((1/a)\arctan(u/a)\).
• The integral \(-\int \frac{u}{u^2 + 9} du\) simplifies using properties of natural logs: it becomes \(-\frac{1}{2}\ln|u^2 + 9|\).
• The second one, \(\int \frac{1}{u^2 + 9} du\), directly uses the inverse tangent formula: \(\frac{1}{3}\arctan\left(\frac{u}{3}\right)\).

These techniques, involving trigonometric and logarithmic functions, facilitate the calculation of integrals that involve squares and can otherwise lead to complicated results.