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Magazine Chapter 7: Problem 71
Evaluate the integral. \( \displaystyle \int \frac{e^{2x}}{1 + e^x}\ dx \)
Short Answer
Expert verified
The integral is \( e^x - \ln|1+e^x| + C \).
Step by step solution
01
Simplify the integral
To evaluate the integral \( \int \frac{e^{2x}}{1 + e^x}\ dx \), we start by simplifying the integrand. Notice that \( e^{2x} = (e^x)^2 \). This allows us to rewrite the integrand as \( \int \frac{(e^x)^2}{1+e^x}\ dx \).
02
Use substitution
Use the substitution method for integration. Let \( u = e^x \), which implies \( du = e^x\ dx \) or \( dx = \frac{du}{u} \). Under this substitution, our integral becomes \( \int \frac{u^2}{1+u} \cdot \frac{du}{u} \).
03
Simplify after substitution
Simplify the integral obtained after substitution. The expression \( \frac{u^2}{1+u} \cdot \frac{1}{u} \) simplifies to \( \frac{u}{1+u} \). Thus, our integral becomes \( \int \frac{u}{1+u}\ du \).
04
Perform polynomial long division
Since the degree of the numerator \( u \) is equal to the degree of the denominator \( 1+u \), perform polynomial long division. Divide \( u \) by \( 1+u \) to get \( 1 - \frac{1}{1+u} \). This means our integral is \( \int \left( 1 - \frac{1}{1+u} \right) du \).
05
Integrate term by term
The integral \( \int \left( 1 - \frac{1}{1+u} \right) du \) can be broken down as \( \int 1\ du - \int \frac{1}{1+u}\ du \). Integrating each term separately, the integral \( \int 1\ du \) is \( u \), and the integral \( \int \frac{1}{1+u}\ du \) is \( \ln|1+u| \).
06
Combine results and substitute back
Combine the integrated terms to get \( u - \ln|1+u| + C \), where \( C \) is the integration constant. Substitute \( u = e^x \) back into the equation to get \( e^x - \ln|1+e^x| + C \).
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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 technique for simplifying integrals by changing variables. It's especially useful when you encounter complex expressions. The idea is to substitute a part of the integrand with a new variable to make the integral easier to solve.
- Identify a substitution: Look for a function inside the integral that, if replaced, simplifies the expression.
- Express the differential: If you choose a function \( u = g(x) \), then find \( du = g'(x) dx \). This connects the old variable to the new one.
- Substitute and simplify: Replace all \( x \) terms with \( u \) terms, including transforming \( dx \) into \( du \). Once substituted, the integrand should be simpler.
- Integrate with respect to \( u \): Carry out the integration as you normally would with this new variable.
- Re-substitute: After integration, replace \( u \) back with the original function \( g(x) \) to get your final answer in terms of \( x \).
Polynomial long division
Polynomial long division is a process to divide two polynomials, similar to how you divide numbers. It's useful when integrands involve polynomial expressions in fractions. Here is a brief how-to:
- Align the terms: Write the terms of both the dividend and divisor in descending order of degrees.
- Divide the leading term: Divide the leading term of the dividend by the leading term of the divisor. This gives the first term of the quotient.
- Multiply and subtract: Multiply the whole divisor by this first term and subtract the result from the dividend. Note down the remaining polynomial.
- Repeat: Continue this process with the new polynomial until the degree of the remainder is less than the degree of the divisor.
Exponential functions
Exponential functions, like \( e^x \), are fundamental in mathematics due to their unique properties, particularly their constant rate of growth. They are defined as \( f(x) = a^x \) where \( a \) is a constant, and usually \( e \) is used, the base of natural logarithms. Many real-world phenomena such as population growth and radioactive decay can be modeled using exponential functions.
- Growth and decay: Exponential functions model systems that increase or decrease at a rate proportional to their current value.
- Derivative property: The function \( e^x \) is unique because its derivative is itself: \( \frac{d}{dx}e^x = e^x \).
- Integral property: Similarly, the integral of \( e^x \) is \( e^x + C \), illustrating how seamlessly exponential functions integrate.
Natural logarithm
The natural logarithm \( \ln(x) \) is the inverse function of an exponential function with base \( e \), and it has many important properties in calculus. It's frequently encountered in integration, particularly with fraction integrands.
- Inverse of exponential: If \( y = e^x \), then \( x = \ln(y) \). This relationship is central when solving equations involving \( e^x \).
- Properties: \( \ln(ab) = \ln(a) + \ln(b) \), \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \), and \( \ln(a^b) = b\ln(a) \).
- Integral of reciprocal: The integral \( \int \frac{1}{x}\, dx \) is \( \ln|x| + C \). This property makes logarithms useful in integrals.
