Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-26, use the method of completing the square, along with a trigonometric substitution if needed, to evaluate each integral. 17\. \(\int \frac{d x}{\sqrt{x^{2}+2 x+5}}\)

Step-by-Step Solution

Answer

\( \int \frac{d x}{\sqrt{x^{2}+2 x+5}} = \ln \left| \frac{x+1 + \sqrt{x^2 + 2x + 5}}{2} \right| + C \)

Step 1: Complete the square

Before evaluating the integral, we complete the square for the expression inside the square root in the denominator: \[x^2 + 2x + 5\]We rewrite it as:\[x^2 + 2x + 1 + 4 = (x + 1)^2 + 4\]

Step 2: Setup trigonometric substitution

Using the identity from completing the square, \[(x+1)^2 + 4\]we let: \[x + 1 = 2 \tan(\theta)\]Then, \[dx = 2 \sec^2(\theta) d\theta\] and the original integral becomes:\[-\int \frac{2 \sec^2(\theta) d\theta}{\sqrt{4\tan^2(\theta) + 4}} = \int \frac{2 \sec^2(\theta) d\theta}{2 \sec(\theta)}\]

Step 3: Simplify the integral

The integral simplifies further:\[\int \sec(\theta) d\theta\]This simplifies to:\[\ln |\sec(\theta) + \tan(\theta)| + C\]

Step 4: Substitute back in terms of x

Recall the substitution: \[x + 1 = 2 \tan (\theta)\]Thus,\[\tan(\theta) = \frac{x+1}{2}\]So,\[\sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + \left(\frac{x+1}{2}\right)^2} = \frac{\sqrt{x^2 + 2x + 5}}{2}\]Substituting these back into the expression gives us:\[\ln \left| \frac{x+1}{2} + \frac{\sqrt{x^2 + 2x + 5}}{2} \right| + C \]

Step 5: Simplify the final expression

Finally, combine fractions inside the logarithm:\[\ln \left| \frac{x+1 + \sqrt{x^2 + 2x + 5}}{2} \right| + C\]This is the simplified form of the integral after back-substitution.

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic expression of the form \(ax^2 + bx + c\) into a perfect square plus a constant. By doing this, integration involving square roots becomes easier, especially when preparing for trigonometric substitution.
To complete the square for the expression \(x^2 + 2x + 5\), we observe that the coefficient of the linear term is 2. We divide this by 2 and square the result, giving us \(1\). We add and subtract this value inside the expression, allowing us to rewrite it as a perfect square:

• Original expression: \(x^2 + 2x + 5\)
• Add and subtract \(1\): \(x^2 + 2x + 1 + 4\)
• Perfect square form: \((x+1)^2 + 4\)

This transformed expression simplifies the process of substituting with trigonometric identities later on in calculus problems.
In summary, completing the square helps us by setting up the quadratic in a form where trigonometric identities can easily be applied, simplifying the integration process.

Definite and Indefinite Integrals

Integrals are a cornerstone of calculus, allowing us to calculate areas under curves and solve differential equations. They come in two main types: definite and indefinite integrals.
Indefinite integrals, like the one in our problem \(\int \frac{dx}{\sqrt{x^2+2x+5}}\), represent a family of functions, expressed with a constant \(C\) because of integration’s nature of reversing differentiation. The solution provides an antiderivative, such as \(\ln |\frac{x+1+\sqrt{x^2+2x+5}}{2}| + C\).
Definite integrals, on the other hand, provide a numeric result by evaluating the area under the curve between two specified limits \([a, b]\). The solution would typically involve substituting these limits into the antiderivative and finding the difference, rather than involving \(C\).
In our context, we are dealing with an indefinite integral, aiming to find the general form of the antiderivative first. The indefinite characteristic also allows flexibility for solving related problems involving given conditions or constraints.

Calculus Problem-Solving

Calculus problem-solving often requires a strategic approach to tackle integrals involving complex expressions. When faced with an integral like \(\int \frac{dx}{\sqrt{x^2+2x+5}}\), techniques like completing the square and trigonometric substitution are invaluable.
First, by completing the square, we turned the inside of the square root into \((x+1)^2 + 4\). This move simplifies the expression and sets the stage for the next technique: trigonometric substitution.

• Identify the substitution: For \((x+1)^2 + 4\), use \(x+1 = 2\tan(\theta)\).
• Transform the differential: \(dx = 2\sec^2(\theta) \, d\theta\).
• Simplify the integral: Transform and solve \(\int \sec(\theta) \, d\theta\).

These steps change the problem into a familiar form that is easier to integrate. Finally, back-substitute using \(\theta = \tan^{-1}((x+1)/2)\) to express the solution in terms of \(x\).
This mix of algebraic manipulation and trigonometric identities is crucial for handling integrals involving more complex algebraic expressions. By mastering these methods, calculus problems become less daunting and more structured.