Question

Evaluate the following integrals. Include absolute values only when needed. $$\int \frac{e^{2 x}}{\left(e^{2 x}+1\right)^{2}} d x$$

Short answer

Answer: The result of the integral is (1/2)(ln|e^(2x) + 1| - ln|e^(2x)|) + C, where C is the constant of integration.

Step-by-step solution

Choose a substitution

Let's make the substitution \(u = e^{2x}+1\). This will simplify the denominator and make the integral easier to evaluate.

Differentiate the substitution with respect to x

We need to calculate \(\frac{du}{dx}\), then we'll be able to replace \(dx\) in the integral: $$\frac{d u}{d x}=\frac{d\left(e^{2 x}+1\right)}{d x}=2e^{2 x}$$

Solve for dx

Now, let's solve for \(dx\) in terms of \(du\): $$d x=\frac{d u}{2 e^{2 x}}$$

Replace the function in the integral with the substitution

Now, let's rewrite the integral in terms of \(u\): $$\int \frac{e^{2 x}}{\left(e^{2 x}+1\right)^{2}} d x = \int \frac{1}{u^{2}} \cdot \frac{d u}{2 e^{2 x}}$$ But we can also note that \(u = e^{2x} + 1 \implies e^{2x} = u - 1\), so we can do one more substitution into the integral: $$\int \frac{1}{u^{2}} \cdot \frac{d u}{2 (u-1)}$$

Simplify the integral

The integral now becomes: $$\frac{1}{2} \int \frac{1}{u(u - 1)} d u$$

Perform partial fraction decomposition

To evaluate this integral, we can use partial fraction decomposition: $$\frac{1}{u(u - 1)} = \frac{A}{u} + \frac{B}{u-1}$$ Solving for \(A\) and \(B\), we find: $$A(u-1) + B(u) = 1$$ $$A = 1$$ $$B = -1$$

Evaluate the integral

With the partial fractions found, we can rewrite the integral: $$\frac{1}{2} \int \left( \frac{1}{u} - \frac{1}{u-1} \right) d u$$ Now, integrate with respect to \(u\): $$\frac{1}{2} \left( \int \frac{1}{u} d u - \int \frac{1}{u-1} d u \right) = \frac{1}{2}(\ln |u| - \ln |u-1|) + C$$

Replace the original function

Finally, let's replace \(u\) back with the original function: $$\frac{1}{2}(\ln |e^{2x} + 1| - \ln |e^{2x}|) + C$$ This is the final answer: $$\boxed{\frac{1}{2}(\ln |e^{2x} + 1| - \ln |e^{2x}|) + C}$$

Key concepts

U-Substitution

Understanding u-substitution is essential when tackling complex integrals. It is a method used to simplify an integral, making it easier to solve. The process involves substituting a part of the integrand (the function to be integrated) with a new variable, typically denoted as u. This is particularly useful when the derivative of u is present elsewhere in the integrand.

For example, if we have an exponential function in the denominator, like in the given exercise \( \int \frac{e^{2x}}{(e^{2x}+1)^{2}} dx \), we can let u = e^{2x}+1 to simplify the expression. After finding \( du \), which is the derivative of u with respect to x, we then rewrite dx in terms of du. This substitution streamlines the integrand, making it more manageable to integrate.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to integrate. When faced with an integral of a fraction where both the numerator and the denominator are polynomials, and the degree of the numerator is less than the degree of the denominator, partial fractions can be a powerful tool.

In the given exercise, after performing u-substitution, we arrived at \( \int \frac{1}{u(u-1)} du \). To integrate this, we express it as the sum of two simpler fractions, \( \frac{A}{u} + \frac{B}{u-1} \). By solving for constants A and B, we simplify the integral into parts that are easier to manage, such as the natural logarithm functions, which brings us to our next topic.

Indefinite Integrals

An indefinite integral, also known as an antiderivative, represents a family of functions whose derivative is the original function we wish to integrate. Unlike a definite integral, an indefinite integral does not evaluate the function over a specific interval; rather, it finds the general form of what function, when differentiated, would give us the integrand.

In the exercise, we perform an indefinite integration of the decomposed fractions. It is important to remember that since the antiderivative is not unique (adding any constant yields another antiderivative) we must include an arbitrary constant C at the end of our solution: \( + C \). This reflects all possible antiderivatives of the integrand.

Natural Logarithm Properties

Natural logarithm properties are indispensable when integrating functions that involve logarithms. The natural logarithm, denoted as \( \ln(x) \), has a base of e, where e is Euler's number, approximately equal to 2.71828. One important property used in integration is \( \ln |ab| = \ln |a| + \ln |b| \), and conversely, \( \ln |\frac{a}{b}| = \ln |a| - \ln |b| \).

In the context of the integral in our exercise, after applying partial fraction decomposition, we obtain the logarithms of two different expressions. By exploiting the properties of logarithms, we are able to integrate and find the difference of two natural logarithms, leading to our final answer in terms of x.