Problem 58 - Calculus for Biology and Medicine

Question

Complete the square in the denominator and evaluate the integral. $$ \int \frac{1}{x^{2}+2 x+5} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C\).

Step 1: Identify the Quadratic Expression

The expression in the denominator is a quadratic: \(x^2 + 2x + 5\). Our goal is to rewrite it by completing the square.

Step 2: Complete the Square in the Quadratic

To complete the square, start with the expression \(x^2 + 2x\). Add and subtract the square of half the coefficient of \(x\), which is \(1\): \(x^2 + 2x + 1 - 1 + 5\). This simplifies to \((x+1)^2 + 4\).

Step 3: Rewrite the Integral

Substitute the completed square form back into the integral: \(\int \frac{1}{(x+1)^2 + 4} \, dx\).

Step 4: Substitute for Simplicity

Let \(u = x+1\), so \(x = u-1\) and \(du = dx\). The integral becomes \(\int \frac{1}{u^2 + 4} \, du\).

Step 5: Recognize the Standard Form

The integral is of the form \(\int \frac{1}{u^2 + a^2} \, du\), with \(a^2 = 4\) or \(a = 2\). The antiderivative is \(\frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C\).

Step 6: Apply the Antiderivative Formula

Using the formula, the integral is \(\frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) + C\).

Step 7: Substitute Back to Original Variable

Replace \(u\) with \(x+1\) to get the final solution: \(\frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C\).

Key Concepts

Completing the Square

Completing the square is a powerful algebraic technique mainly used in integration to simplify complex expressions. When you encounter a quadratic expression, it can be rewritten in a form that is squareroot friendly.
For the quadratic expression in the exercise, \(x^2 + 2x + 5\), you aim to express it as a perfect square plus a constant.

Here’s a simple guide to practice:

• Isolate the quadratic and linear terms: Take \(x^2 + 2x\) separately.
• Add and subtract \((\frac{b}{2})^2\): Where \(b\) is the coefficient of \(x\). Here, \(2\), so \((\frac{2}{2})^2 = 1\). You add and also subtract this value to keep the expression unchanged: \(x^2 + 2x + 1 - 1 + 5\).
• Rewrite: Now, it is easier to see \((x+1)^2 + 4\).

This makes using other integration techniques more straightforward, such as trigonometric substitution.

Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate integrals that involve square roots or quadratic expressions. After completing the square, it's common to implement this method to transform the integral
into something that matches a trigonometric identity, simplifying the integration process.

Here’s how you utilize trigonometric substitution based on the completed square:

• With the expression \((x+1)^2 + 4\), or written as \(u^2 + 4\) after substitution, recognize it is in the form \(u^2 + a^2\).
• Substitute: Use \(u = x+1\) and choose \(u = a \tan(\theta)\) where \(a = 2\) to fit the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\).
• Simplify: The differential \(du = a \sec^2(\theta) d\theta\) can replace \(du\), facilitating the integration of trigonometric functions.

Using trigonometric identities like these simplifies the antiderivative finding process, making it possible to arrive at the exact result quickly.

Indefinite Integral

An indefinite integral, sometimes referred to as an antiderivative, represents a fundamental concept in calculus. Unlike a definite integral, which calculates a precise area under the curve, an indefinite integral offers a family of functions representing all possible antiderivatives of the original function.

To unpack the process with the example

• After completing the square and substituting, we encounter the standard formula: \( \int \frac{1}{u^2 + a^2} \, du\).
• This aligns with the recognizable antiderivative formula: \( \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C\), where \(C\) indicates a constant of integration.
• The substitution shifts back after integration to keep expressions in terms of the original variable \(x\).

Mastering indefinite integrals involves recognizing form patterns you translate into known antiderivative formulas. This is essential for simplifying expressions and finding solutions to calculus problems efficiently.