Problem 80 - Thomas Calculus

Question

Evaluate each integral in Exercises \(71-82\) by using any technique you think is appropriate. $$ \int \frac{d x}{(2 x+1) \sqrt{4 x^{2}+4 x}} $$

Step-by-Step Solution

Answer

\( \frac{1}{2} \sqrt{x(x+1)} + C \)

Step 1: Simplify the Expression Inside the Integral

The given integral is \( \int \frac{d x}{(2 x+1) \sqrt{4 x^{2}+4 x}} \). Begin by simplifying the square root expression: \( \sqrt{4x^2 + 4x} = \sqrt{4(x^2 + x)} = \sqrt{4} \cdot \sqrt{x(x+1)} = 2\sqrt{x(x+1)} \). Therefore, the integral becomes \( \int \frac{d x}{(2 x+1) \cdot 2\sqrt{x(x+1)}} \).

Step 2: Simplify the Integral

We can simplify the integral further by factoring constant terms outside the integral: \( \frac{1}{2} \int \frac{d x}{(2 x+1) \sqrt{x(x+1)}} \).

Step 3: Use Substitution Technique

Use the substitution \( u = \sqrt{x(x+1)} \) to simplify the integral. Compute \( du \) by first taking the derivative: \( u = \sqrt{x(x+1)} = \sqrt{x^2 + x} \), giving us \( du = \frac{x + \frac{1}{2}}{\sqrt{x^2 + x}} \, dx \). Solve for \( dx \): \( dx = \frac{2\sqrt{x^2 + x}}{2x+1} \, du \). Substitute \( dx \) back into the integral.

Step 4: Perform the Integral Substitution

The integral becomes \( \frac{1}{2} \int \frac{\frac{2\sqrt{x^2+x}}{2x+1}\, du}{(2 x+1) \cdot \sqrt{x(x+1)}} \). The terms simplify to \( \frac{1}{2} \int du \), which is \( \frac{1}{2} u + C \) after integration.

Step 5: Back-Substitute \( u \)

Substitute back the original function for \( u \). Recall that \( u = \sqrt{x(x+1)} \), so the solution is \( \frac{1}{2} \sqrt{x(x+1)} + C \).

Key Concepts

Integration Techniques

In the world of integral calculus, there are several methods you can use to evaluate integrals. These methods help transform a complex integral into a simpler one, or even an integral you have encountered before. Common techniques include:

• Substitution Method: Changing variables to make an integral easier to solve.
• Integration by Parts: Useful when dealing with products of functions.
• Partial Fraction Decomposition: Breaking fractions into simpler parts.
• Trigonometric Substitution: Replacing variables with trigonometric functions.

Each technique has its strengths and appropriate scenarios for use. For example, substitution works wonders when you spot a derivative within the integral. By practicing, you'll get better at identifying the best technique to use.

Substitution Method

This technique is incredibly powerful when the integral contains a function and its derivative. The idea is to simplify the integral by introducing a new variable. Let's look at how this works:

• Identify the inner function: In integrals with compositions, find part of the integrand that can be substituted with a single variable.
• Change of variable: Substitute with a new variable (often denoted as 'u'). For instance, if you have a term like \(x^2 + x\), let \(u = x^2 + x\).
• Find differntial \(du\): Differentiate the expression to express \(dx\) in terms of \(du\). For example, if \(u = x^2 + x\), then \(du = (2x + 1) \, dx\).
• Integrate: With the new variable, the integral becomes simpler. You integrate with respect to the new variable.
• Re-substitute: Finally, replace back the original variables to express the integral in its initial terms.

This method is especially useful when dealing with radicals or when a part of the integrand behaves like an inner derivative.

Definite and Indefinite Integrals

Integrals come in two main types: definite and indefinite. Understanding the difference between them is crucial for solving calculus problems effectively.

• Indefinite Integrals produce a family of functions and include a constant of integration \(C\). They take the form \(\int f(x) \, dx = F(x) + C\).
• Definite Integrals give a numerical result by evaluating the integrant over a specific interval \([a, b]\). They are expressed as \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\).

Indefinite integrals focus on finding the antiderivative of a function, while definite integrals involve the area under a curve between two points. When working with integrals, it is important to identify if they are definite or indefinite to apply the correct solution approach.

Integral Simplification

Simplifying an integral can sometimes be the key to evaluating it. By reducing complexity, you make the integral easier to solve through straightforward techniques. Here are some strategies:

• Factor and simplify: Look for factors in both the numerator and denominator that might cancel each other out.
• Use trigonometric identities: Sometimes, trigonometric functions can be transformed into simpler terms.
• Simplify radicals: Techniques like multiplying by conjugates can simplify terms under a square root.
• Decompose fractions: Breaking complex rational functions into simpler parts can make integration straightforward.

These simplifications often reduce the need for more complicated integral techniques. They transform the integral into a form much easier to work with, allowing the basic integration skills to shine.