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Magazine Chapter 8: Problem 2
Use the table of integrals at the back of the book to evaluate the integrals. $$\int \frac{d x}{x \sqrt{x+4}}$$
Short Answer
Expert verified
\( \frac{1}{2} \ln \left| \frac{\sqrt{x+4} - 2}{\sqrt{x+4} + 2} \right| + C \)
Step by step solution
01
Identify the Integral Form
Look through the table of integrals to find a form that matches the integrand \( \frac{1}{x \sqrt{x+4}} \). We need an integral that resembles this expression.
02
Use Substitution
Choose an appropriate substitution. Let \( u = \sqrt{x+4} \), then \( u^2 = x + 4 \), which implies \( x = u^2 - 4 \). Differentiating both sides, we get \( dx = 2u \, du \).
03
Rewrite the Integral
Substitute \( x = u^2 - 4 \) and \( dx = 2u \, du \) into the integral:\[\int \frac{1}{(u^2 - 4)u} \cdot 2u \, du = 2 \int \frac{u}{(u^2 - 4)u} \, du = 2 \int \frac{1}{u^2 - 4} \, du\]
04
Simplify the Integral Expression
The integral simplifies to \( 2 \int \frac{1}{u^2 - 4} \, du \). This is a standard form that can typically be solved by recognizing it as a partial fraction decomposition problem.
05
Evaluate Using Partial Fractions
Express \( \frac{1}{u^2 - 4} = \frac{1}{(u-2)(u+2)} \). Use partial fraction decomposition: \[\frac{1}{u^2 - 4} = \frac{A}{u - 2} + \frac{B}{u + 2}\]Solving for \( A \) and \( B \), you find that \( A = \frac{1}{4} \) and \( B = -\frac{1}{4} \).
06
Integrate the Partial Fractions
Now integrate each part:\[2 \left( \int \frac{1}{4(u-2)} \, du - \int \frac{1}{4(u+2)} \, du \right)\]This results in:\[\frac{1}{2} \left( \ln|u-2| - \ln|u+2| \right) + C = \frac{1}{2} \ln \left| \frac{u-2}{u+2} \right| + C\]
07
Back-Substitute to Original Variable
Recall \( u = \sqrt{x+4} \). Substitute back to get:\[\frac{1}{2} \ln \left| \frac{\sqrt{x+4} - 2}{\sqrt{x+4} + 2} \right| + C\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful tool in integral calculus that helps simplify complex expressions, making them easier to evaluate. This technique involves substituting a part of the integral with a new variable, usually denoted as \( u \). The idea is to transform a challenging integral into a simpler form.
First, identify a function within the integral which, when substituted with a new variable, simplifies the integrand. In this exercise, we chose \( u = \sqrt{x+4} \).
First, identify a function within the integral which, when substituted with a new variable, simplifies the integrand. In this exercise, we chose \( u = \sqrt{x+4} \).
- After substitution, express the differential in terms of \( du \). Here, differentiating \( u = \sqrt{x+4} \), we find \( dx = 2u \, du \).
- Replace all instances of the original variable in the integrand with the expressions involving \( u \) and \( du \).
Partial Fraction Decomposition
Once you have simplified an integral using substitution, the next step often involves partial fraction decomposition. This technique is used to break down a complex rational expression into simpler fractions that are easier to integrate.
For our integral, after substitution, we had \( \int \frac{1}{u^2 - 4} \, du \). Recognizing \( u^2 - 4 \) as a difference of squares, we can express it as \( (u-2)(u+2) \).
For our integral, after substitution, we had \( \int \frac{1}{u^2 - 4} \, du \). Recognizing \( u^2 - 4 \) as a difference of squares, we can express it as \( (u-2)(u+2) \).
- Solve for constants \( A \) and \( B \) in the equation \( \frac{1}{u^2 - 4} = \frac{A}{u - 2} + \frac{B}{u + 2} \).
- This requires setting up an equation by equating coefficients, leading to \( A = \frac{1}{4} \) and \( B = -\frac{1}{4} \).
Definite Integrals
Definite integrals are used to compute the exact area under a curve defined by a function over a specific interval. While our task here involves an indefinite integral, understanding definite integrals is crucial.
A definite integral encompasses bounds, such as \( \int_{a}^{b} f(x) \, dx \). It provides a number representing the total accumulated quantity over that interval.
A definite integral encompasses bounds, such as \( \int_{a}^{b} f(x) \, dx \). It provides a number representing the total accumulated quantity over that interval.
- Using the Fundamental Theorem of Calculus, we connect definite and indefinite integrals by evaluating the antiderivative at the upper and lower bounds \( F(b) - F(a) \).
- Substitution and partial fractions can be applied similarly as in indefinite integrals, but pay attention to the bounds during back-substitution.
