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Magazine Chapter 5: Problem 9
Evaluate the integrals using the indicated substitutions. (a) \(\int \frac{d x}{x \ln x} ; u=\ln x\) (b) \(\int e^{-5 x} d x ; u=-5 x\)
Short Answer
Expert verified
(a) \( \ln |\ln x| + C \), (b) \( -\frac{1}{5}e^{-5x} + C \)
Step by step solution
01
Substitution for Integral (a)
Start by substituting the given variable: \( u = \ln x \). Then, the derivative \( \frac{du}{dx} = \frac{1}{x} \) gives us the differential \( du = \frac{1}{x} dx \), or equivalently, \( dx = x \, du \).
02
Rewrite Integral (a)
Now substitute \( u = \ln x \) and the expression for \( dx \) into the integral: \[ \int \frac{dx}{x \ln x} = \int \frac{1}{x} \cdot \frac{x \, du}{u} = \int \frac{du}{u}. \]
03
Evaluate Integral (a)
The integral \( \int \frac{du}{u} \) is a standard logarithmic integral, which evaluates to \( \ln |u| + C \), where \( C \) is the constant of integration. Since \( u = \ln x \), we substitute back to get \( \ln |\ln x| + C \).
04
Substitution for Integral (b)
Use the substitution \( u = -5x \). The derivative \( \frac{du}{dx} = -5 \) gives us \( du = -5 \, dx \), so \( dx = -\frac{1}{5} du \).
05
Rewrite Integral (b)
Substitute \( u = -5x \) and the expression for \( dx \) into the integral: \[ \int e^{-5x} \, dx = \int e^{u} \left(-\frac{1}{5}\right) \, du = -\frac{1}{5} \int e^{u} \, du. \]
06
Evaluate Integral (b)
The integral \( \int e^u \, du \) evaluates to \( e^u + C \). Multiplying by \(-\frac{1}{5}\), we have \( -\frac{1}{5}(e^u) + C = -\frac{1}{5}e^{-5x} + C \) after substituting \( u = -5x \).
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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 in calculus is a pivotal technique used to simplify the process of finding integrals. It is similar to the process of "changing variables," which can often make a complex integral much more manageable. Here鈥檚 how it works:
- Identify the substitution: Begin by spotting a part of the integral that seems complex or cumbersome. By substituting this part with a new variable, typically like "u," you aim to simplify the integral. For example, in integral (a) from the exercise, we chose the substitution \(u = \ln x\).
- Determine \(du\): Derive \(du\) in terms of \(dx\). If \(u = \ln x\), then the derivative \(\frac{du}{dx} = \frac{1}{x}\), and subsequently, \(du = \frac{1}{x}dx\).
- Replace and Solve: Substitute everything in the integral to terms of \(u\) and \(du\). This turns the original integral into a simpler one that is easier to evaluate. For instance, \(\int \frac{1}{x \ln x} dx\) becomes \(\int \frac{du}{u}\).
Logarithmic Integrals
Logarithmic integrals appear frequently in calculus, often described by the expression \(\int \frac{1}{x} dx\). In our exercise, the transformed integral \(\int \frac{du}{u}\) is a classic logarithmic form.
- Evaluation: The integral \(\int \frac{du}{u}\) leads to the result \(\ln |u| + C\). Here, the constant \(C\) is crucial as it accounts for all the specific solutions to the indefinite integral.
- Substituting Back: After integrating, remember to substitute back the expression you initially replaced with \(u\). Since \(u = \ln x\), the final solution becomes \(\ln |\ln x| + C\).
Exponential Integrals
Exponential integrals involve functions with the base of the natural logarithm e, such as \(e^x\). They often require clever manipulation to solve effectively, often done using substitution.
- General Form: The basic exponential integral is \(\int e^x dx\), which results in \(e^x + C\). However, integrals can become more complex when dealing with functions like \(\int e^{-5x} dx\).
- Using Substitution: The substitution method is invaluable here. For example, \(u = -5x\) simplifies the integral \(\int e^{-5x} dx\) to \(-\frac{1}{5} \int e^u \, du\), instantly solvable to \(-\frac{1}{5} e^u + C\).
- Substitute Back: Transition back to the original variable by replacing \(u\) with \(-5x\), resulting in \(-\frac{1}{5} e^{-5x} + C\), which is the evaluated function.
