Question

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{d x}{\left(e^{x}+e^{-x}\right)^{2}}$$

Short answer

Based on the given step-by-step solution, the integral \(\int \frac{d x}{\left(e^{x}+e^{-x}\right)^{2}}\) evaluates to \(-\frac{1}{4\left(e^{2x} + 2 + e^{-2x}\right)} + C\).

Step-by-step solution

Perform Change of Variables

Let's make the change of variables: $$u = e^x, du = e^x dx.$$ We also need to find the expression for \(e^{-x}\). As \(e^{-x} = \frac{1}{e^x}\), we can substitute \(u\) to get \(e^{-x} = \frac{1}{u}\). Now, let's substitute these values in the given integral. $$\int \frac{d x}{\left(u+\frac{1}{u}\right)^{2}}\times\frac{du}{u}$$

Simplify the Integral

Now, let's simplify the integral by multiplying both the numerator and denominator of the fraction by \(u^2\) to get rid of the fractions inside the integral. $$\int \frac{u^{2}}{\left(u^2 + 2 + u^{-2}\right)^2} du$$

Integrate

Now, let's use the substitution method to evaluate the integral. Let $$v = u^2 + 2 + \frac{1}{u^2}, dv = 4u du.$$ Then, the integral becomes: $$\frac{1}{4}\int \frac{4u^3}{v^2} dv$$ Using substitution, we get: $$\frac{1}{4}\int \frac{1}{v^2} dv$$ Now, let's integrate this expression: $$\frac{1}{4} \left(-\frac{1}{v}\right) + C$$

Substitute Values back to Original Variables

Substituting back the values for \(v\), we get: $$-\frac{1}{4\left(u^2 + 2 + \frac{1}{u^2}\right)} + C$$ Finally, we SUBSTITUTE back the original variable \(x\): $$-\frac{1}{4\left(e^{2x} + 2 + e^{-2x}\right)} + C$$ So, the result of the integral is: $$\int \frac{d x}{\left(e^{x}+e^{-x}\right)^{2}} = -\frac{1}{4\left(e^{2x} + 2 + e^{-2x}\right)} + C$$

Key concepts

Change of Variables

In calculus, a change of variables is a handy technique to simplify complex integrals by introducing a new variable. The goal here is to make the integral easier to solve. For instance, when dealing with exponential functions, substituting a term can often transform the integral into a more workable form.

In this example, we choose to let the new variable be:

• \( u = e^x \), which leads to \( du = e^x dx \) as the differential form.
• Also, note that \( e^{-x} = \frac{1}{u} \), this makes substitution straightforward.

Substituting these into the integral, we replace every instance of the original variables with the new variable \( u \). This results in transforming the complex expression into a simpler one with respect to \( du \). This approach not only simplifies computation but also helps in seeing the integral's structure more clearly.

The logic follows as substituting into the integral \( \int \frac{d x}{\left(e^{x}+e^{-x}\right)^{2}} \), helps us avoid cumbersome expressions integrating directly with \( x \). Keeping a clear track of variable transformations ensures that no step is missed and reversed correctly at the end.

Integration

Integration is about finding the area under a curve, with indefinite integrals representing families of functions differing by a constant. The integration process begins after setting the integral with the new variable, which in this case presents an expression more amenable to calculus techniques.

Following change of variables, the integral simplifies and allows multiplication to clear fractions, paving the path to an easier form, such as:

• \( \int \frac{u^2}{(u^2 + 2 + u^{-2})^2} du \)

This integral again suggests setting up another substitution since it is still in a format that's not instantly integrable. We introduce:

• \( v = u^2 + 2 + \frac{1}{u^2} \), simplifying the integral into terms of \( dv \).

A well-crafted choice can often turn a seemingly difficult problem into a straightforward task. Integration using substitution works best when each step logically leads to a downstream simplification, guiding the correct antiderivative.

Calculus Techniques

Calculus techniques involve a series of strategies in both differentiation and integration that ensure complicated problems are broken down into simpler, solvable pieces. One such strategy deployed here involves substitution not once but twice, showing the flexibility and power of substitution.

Substitution is just one tool among many; it is vital to appreciate where the method fits appropriately. After postulating \( v = u^2 + 2 + \frac{1}{u^2} \), the resultant integral \( \int \frac{1}{v^2} dv \) becomes manageable.

We find:

• \( \frac{1}{4} \left(-\frac{1}{v}\right) + C \) serves as our antiderivative in terms of the substituted variable.

Finally, reverse the variable changes to return to the original function form! While substitution is central here, keeping a broad array of calculus techniques in one's toolbox aids problem-solving in calculus. Techniques such as partial fractions, integration by parts, and series expansion, all serve their niche but can be bolstered greatly by understanding and applying substitution effectively.