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Magazine Chapter 6: Problem 38
Find each integral by integration by parts or a substitution, as appropriate. a. \(\int \sqrt{\ln x} \frac{1}{x} d x\) b. \(\int x^{2} e^{x^{3}} d x\) c. \(\int x^{7} \ln 3 x d x\) d. \(\int x e^{4 x} d x\)
Short Answer
Expert verified
a. \(\frac{2}{3}(\ln x)^{3/2} + C\); b. \(\frac{1}{3} e^{x^3} + C\); c. \(\frac{x^8 \ln 3x}{8} - \frac{x^8}{192} + C\); d. \(\frac{x}{4} e^{4x} - \frac{1}{16} e^{4x} + C\).
Step by step solution
01
Step 1a: Identify the Substitution for Part A
Let's consider the integral \( \int \sqrt{\ln x} \frac{1}{x} \, dx \). We notice the composition \( \ln x \), suggesting a substitution: let \( u = \ln x \), then \( du = \frac{1}{x} \, dx \). This transforms the integral into \( \int \sqrt{u} \, du \).
02
Step 2a: Solve the Integral for Part A
Now, solve \( \int \sqrt{u} \, du \), which can be rewritten as \( \int u^{1/2} \, du \). By using the power rule of integration, we get \( \frac{u^{3/2}}{3/2} + C \) or equivalently \( \frac{2}{3} u^{3/2} + C \). Substitute back \( u = \ln x \) to get \( \frac{2}{3} (\ln x)^{3/2} + C \).
03
Step 1b: Identify the Substitution for Part B
Consider the integral \( \int x^{2} e^{x^{3}} \, dx \). Here, the expression \( x^3 \) inside the exponential suggests the substitution: let \( u = x^3 \), then \( du = 3x^2 \, dx \). Rewrite \( dx \) as \( dx = \frac{du}{3x^2} \). The integral becomes \( \frac{1}{3} \int e^{u} \, du \).
04
Step 2b: Solve the Integral for Part B
Integrate \( \int e^u \, du \), which directly evaluates to \( e^u + C \). Multiplying by \( \frac{1}{3} \), the result is \( \frac{1}{3} e^u + C \). Substitute back \( u = x^3 \) to get \( \frac{1}{3} e^{x^3} + C \).
05
Step 1c: Choose Integration by Parts for Part C
Consider the integral \( \int x^{7} \ln 3x \, dx \). Use integration by parts where \( u = \ln 3x \) and \( dv = x^7 \, dx \). Then \( du = \frac{1}{3x} \, dx \) and \( v = \frac{x^8}{8} \). The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).
06
Step 2c: Perform Integration by Parts for Part C
Apply the formula: \( \int x^{7} \ln 3x \, dx = \frac{x^8 \ln 3x}{8} - \int \frac{x^8}{8} \cdot \frac{1}{3x} \, dx \). Simplify the integrand: \( \frac{x^7}{24} \), leading to \( \int \frac{x^7}{24} \, dx \). Solve to get \( \frac{x^8}{192} + C \). Thus, the integration is: \( \frac{x^8 \ln 3x}{8} - \frac{x^8}{192} + C \).
07
Step 1d: Use Integration by Parts for Part D
Consider the integral \( \int x e^{4x} \, dx \). Choose parts: \( u = x \), \( dv = e^{4x} \, dx \) resulting in \( du = dx \), \( v = \frac{1}{4} e^{4x} \). Use the integration by parts formula \( \int u \, dv = uv - \int v \, du \).
08
Step 2d: Calculate Integration by Parts for Part D
Apply the formula: \( \int x e^{4x} \, dx = \frac{x}{4} e^{4x} - \int \frac{1}{4} e^{4x} \, dx \). Evaluate the remaining integral: \( -\frac{1}{16} e^{4x} + C \). Thus, the solution is \( \frac{x}{4} e^{4x} - \frac{1}{16} e^{4x} + C \).
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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, helping to simplify complex expressions. We often use it when an exponent or a logarithmic function appears inside a larger formula, as seen in Part A and Part B of our exercise.
To apply the Substitution Method:
To apply the Substitution Method:
- Identify a part of the integral that can be simplified by a change of variable. This is usually an inside function of a composite function, like the inner part of a logarithm or a power.
- Introduce a new variable to replace this part. For example, in Part A, set \( u = \ln x \), which makes \( du = \frac{1}{x} \, dx \), turning the integral to \( \int \sqrt{u} \, du \).
- Calculate the derivative of the substitution and adjust \( dx \) accordingly. In Part B, where \( u = x^3 \), we have \( du = 3x^2 \, dx \) and \( dx = \frac{du}{3x^2} \).
Power Rule of Integration
The Power Rule of Integration is a straightforward method to find the integral of a function with a power of \( x \). It is expressed as:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
where \( n eq -1 \). This rule was used in solving Part A. Here's how it works:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
where \( n eq -1 \). This rule was used in solving Part A. Here's how it works:
- Recognize the integral of the form \( \int u^n \, du \), like \( \int \sqrt{u} \, du \) which is \( \int u^{1/2} \, du \) in our example.
- Apply the power rule by increasing the exponent by 1 and dividing by the new exponent. Therefore, \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C \).
- Substitute back the original variable to complete the integration.
Exponential Integration
Exponential Integration involves integrals that include exponential functions, typically \( e^x \). The simplicity of these functions is what makes them beautiful and powerful. For example, in Part B and Part D of our exercise:
To solve an integral like \( \int e^{u} \, du \):
To solve an integral like \( \int e^{u} \, du \):
- Recognize that the integral of an exponential function is straightforward: \( \int e^u \, du = e^u + C \).
- Pay attention to any coefficients or additional parts in the integrand. For Part B, \( x^2 e^{x^3} \): we used substitution to simplify to \( \frac{1}{3} \int e^{u} \, du \).
- Ensure that any constants, like \( \frac{1}{3} \), remain factored throughout the integration.
- After integrating, remember to substitute back to get back to the original variable context.
Logarithmic Integration
Logarithmic Integration often requires the method of integration by parts, especially when a logarithmic function multiplies a polynomial, as shown in Part C.
Essential steps in Logarithmic Integration involve:
Essential steps in Logarithmic Integration involve:
- Identifying parts of the integrand where a logarithm is present. Here, we used \( u = \ln 3x \) and \( dv = x^7 \, dx \).
- Applying Integration by Parts formula: \( \int u \, dv = uv - \int v \, du \).
- Determine \( du \) and \( v \) based on our choices. In this case, \( du = \frac{1}{3x} \, dx \) and \( v = \frac{x^8}{8} \).
- Simplifying the subsequent integrals that arise from the integration by parts formula.
