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91Ó°ÊÓ

Find each indefinite integral. \(\int\left(e^{3 x}-\frac{3}{x}\right) d x\)

Short Answer

Expert verified
The integral is \( \frac{1}{3} e^{3x} - 3 \ln|x| + C \).

Step by step solution

01

Integrate each term separately

Begin by splitting the integral into two separate integrals: \( \int e^{3x} \, dx - \int \frac{3}{x} \, dx \). This allows us to integrate each term individually.
02

Integrate the exponential function

The integral of \( e^{3x} \) can be found by using substitution. Let \( u = 3x \), thus \( du = 3 \, dx \) or \( dx = \frac{1}{3} \, du \). This transforms \( \int e^{3x} \, dx \) to \( \int \frac{1}{3} e^u \, du = \frac{1}{3} e^u + C_1 \). Substitute back \( u = 3x \) to get \( \frac{1}{3} e^{3x} + C_1 \).
03

Integrate the rational function

To integrate \( \int \frac{3}{x} \, dx \), recall that the integral of \( \frac{1}{x} \) is \( \ln|x| \). Therefore, \( \int \frac{3}{x} \, dx = 3 \ln|x| + C_2 \).
04

Combine the results and simplify

Combine the results from Steps 2 and 3: \( \frac{1}{3} e^{3x} - 3 \ln|x| + C \), where \( C \) is the combination of integration constants \( C_1 \) and \( C_2 \). This gives the complete indefinite integral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Function Integration
Integrating exponential functions can sometimes seem tricky, but once you understand a few simple steps, it becomes much easier. When dealing with integrals such as \( \int e^{3x} \, dx \), we use a method called substitution to simplify the process.

Here's how it works:
  • First, identify that the exponent in the function is a linear expression, like \( 3x \). We set \( u = 3x \), which simplifies our process.
  • Next, find the differential \( dx \) in terms of \( du \). For \( u = 3x \), differentiate to get \( du = 3 \cdot dx \), hence \( dx = \frac{1}{3} du \).
  • Substitute these into the integral: \( \int e^{3x} \, dx = \int e^u \frac{1}{3} du \).
The integral \( \int e^u \ du \) simplifies to \( e^u \). So, the integral becomes \( \frac{1}{3} e^u + C_1 \) where \( C_1 \) is a constant of integration. Finally, substitute \( u \) back to get \( \frac{1}{3} e^{3x} + C_1 \). This process allows you to find the indefinite integral of exponential functions effectively.
Rational Function Integration
Rational functions, like \( \frac{3}{x} \), often appear intimidating, but they aren't as complex as they seem. The key is to remember the natural logarithm function. Integrating \( \frac{3}{x} \) involves recalling a basic integral identity.

Here's a concise breakdown:
  • The integral \( \int \frac{1}{x} \, dx \) results in \( \ln|x| \). This is because the derivative of \( \ln|x| \) is \( \frac{1}{x} \).
  • Given \( \int \frac{3}{x} \, dx \), you can pull out the constant (3) outside the integral: \( 3 \int \frac{1}{x} \, dx \).
Once this is done, apply our integral identity: \( 3 \ln|x| + C_2 \), where \( C_2 \) is another constant of integration.

By focusing on core identities like \( \frac{1}{x} \Rightarrow \ln|x| \), rational integrations can be tackled with ease and precision.
Integration by Substitution
The process of integration by substitution is a vital skill in calculus, often used to simplify complex integrals. In our given problem, you may have noticed it was applied to the exponential component of the integral: \( \int e^{3x} \, dx \).

This method follows these basic steps:
  • Choose a substitution that simplifies the integral. Typically, this involves substituting a complicated expression with a simpler variable \( u \), like \( u = 3x \).
  • Differentiate this substitution to find \( du \), which helps in rewriting \( dx \) in terms of \( du \).
  • Replace the original variable and differential with \( u \) and \( du \) in the integral. This often turns the integral into a more straightforward problem.
After integrating with respect to \( u \), always substitute back the original variable to get a solution in terms of the initial variable, which in our exercise was \( x \).

Integration by substitution is akin to undoing the chain rule from differentiation, making it an essential and powerful technique in calculus.

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