Question

Perform the indicated integrations. $$ \int \frac{1}{x^{2}+2 x+5} d x $$

Short answer

\( \frac{1}{2} \tan^{-1} \frac{x+1}{2} + C \)

Step-by-step solution

Identify the form

The given integral is \( \int \frac{1}{x^{2}+2x+5} \, dx \). Notice that the denominator is a quadratic expression, which suggests the use of a completing the square method or trigonometric substitution.

Complete the square

Rewrite the expression in the denominator by completing the square. Start with \( x^{2} + 2x + 5 \). The square can be completed by adding and subtracting \((\frac{2}{2})^2 = 1\), which gives: \((x+1)^2 + 4\). Thus, rewrite the integral as \( \int \frac{1}{(x+1)^2 + 2^2} \, dx \).

Use trigonometric substitution

Recognize that the expression \((x + 1)^2 + 2^2\) resembles the form \(a^2 + u^2\) which is suitable for the arctangent substitution. Let \( u = x + 1 \), so \( du = dx \). The integral becomes \( \int \frac{1}{u^2 + 2^2} \, du \).

Apply standard arctangent integration

The integral \( \int \frac{1}{u^2 + a^2} \, du \) is \( \frac{1}{a} \tan^{-1} \frac{u}{a} + C \). Here, \( a = 2 \). Substitute into the formula to get: \( \frac{1}{2} \tan^{-1} \frac{u}{2} + C \).

Reverse the substitution

Replace \( u \) with \( x+1 \) to revert back to the variable \( x \), giving: \( \frac{1}{2} \tan^{-1} \frac{x+1}{2} + C \). This is the solution to the original integral.

Key concepts

Completing the Square

Completing the square is a method used to simplify quadratic expressions, particularly useful when dealing with integral calculus. It's exactly what the name implies – restructuring a quadratic expression so that it becomes a perfect square trinomial. Let's break it down.

Starting with a general quadratic expression, such as \(x^2 + bx + c\), the goal is to rewrite it in the form \((x-h)^2 + k\). The term \(h\) is determined as \(\frac{b}{2}\), and adding and subtracting \((\frac{b}{2})^2\) helps achieve this structure.

For the example \(x^2+2x+5\), we have:

• The linear coefficient \(b\) is 2, so \(\frac{b}{2} = 1\).
• Add and subtract \(1^2 = 1\), to get: \(x^2 + 2x + 1 - 1 + 5\).
• This rearranges to \((x+1)^2 + 4\).

This transformation helps simplify the integration process by making the expression easier to manipulate or substitute.

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in integration when dealing with certain types of algebraic expressions. It's particularly useful when an integral involves expressions in the form \(a^2 + x^2\), \(a^2 - x^2\), or \(x^2 - a^2\), as these can be connected to trigonometric identities.

In this exercise, after completing the square, we have an expression \((x+1)^2 + 2^2\). This fits well with the \(a^2 + u^2\) form, where trigonometric identities involving tangent can simplify the problem.

Here's how it simplifies:

• Set the substitution \(u = x + 1\), making \(du = dx\).
• The integral becomes \(\int \frac{1}{u^2 + 2^2} \, du\), aligning with the trigonometric identity.

Trigonometric substitution sets a foundation for the next step in integral calculus, which involves recognizing and applying specific integral formulas.

Arctangent Integration

Arctangent integration is a technique for integrating expressions that match the standard form \(\int \frac{1}{u^2 + a^2} \, du\). This form is particularly well-suited for using arctangent functions, a common function in calculus that helps simplify certain types of integrals.

Here's how it works:

• For \(\int \frac{1}{u^2 + 2^2} \, du\), recognize it corresponds to the standard form \(\int \frac{1}{u^2 + a^2} \, du\).
• Using the formula, the result is \(\frac{1}{a} \tan^{-1} \frac{u}{a} + C\), where \(a\) is 2 in this case.
• This results in \(\frac{1}{2} \tan^{-1} \frac{u}{2} + C\).

After applying the formula, simply reverse the substitution from earlier, replacing \(u\) back with the original expression \(x+1\). This yields the final result: \(\frac{1}{2} \tan^{-1} \frac{x+1}{2} + C\), providing the elegant solution to the integral.