Question

Evaluate the following integrals. $$\int \frac{68}{e^{2 x}+2 e^{x}+17} d x$$

Short answer

Question: Evaluate the following integral: $$\int \frac{68}{e^{2x}+2e^x+17} dx$$ Answer: $$\int \frac{68}{e^{2 x}+2 e^{x}+ 17} d x = 4x + 2\ln(e^{2x}+2e^x+17) + C$$

Step-by-step solution

Perform the \(u\)-substitution

Let \(u = e^x\). Hence, \(du = e^x dx\). Notice that \(dx = \frac{du}{u}\). Now, we can rewrite \(e^{2x}\) in terms of \(u\). We have: $$e^{2x} = (e^x)^2 = u^2$$ Our integral is now: $$\int \frac{68}{u^2+2u+17} \frac{du}{u}$$

Simplify the rational function

Now we can simplify the rational function: $$\frac{68}{u(u^2+2u+17)}du$$

Separate into partial fractions

The next step is to separate it into partial fractions: $$\frac{A}{u}+\frac{Bu+C}{u^2+2u+17}$$ We can find the values of \(A\), \(B\), and \(C\) by setting: $$68 = A(u^2+2u+17) + (Bu+C)u$$

Find the values of \(A\), \(B\), and \(C\)

By equating coefficients, we have: \begin{align*} A + B &= 0 \\ 2A + C &= 0 \\ 17A &= 68 \end{align*} Solving this system of equations, we get \(A=4\), \(B=-4\), and \(C=8\). Thus, our partial fractions decomposition is: $$\frac{4}{u}-\frac{4u+8}{u^2+2u+17}$$

Integrate the partial fractions

Now we can integrate the two partial fractions: $$\int \left(\frac{4}{u}-\frac{4u+8}{u^2+2u+17}\right) du$$ The first term is easy to integrate: $$\int \frac{4}{u} du = 4\ln |u| + C_1$$ For the second term, let's use another substitution since the numerator is the derivative of the denominator: $$v = u^2+2u+17 \Rightarrow dv = (2u+2) du$$ Now, we notice that \(\frac{4u+8}{u^2+2u+17}=\frac{2(2u+2)}{v}\), and integrate: $$2\int \frac{dv}{v} = 2\ln |v| + C_2$$

Substitute back

Now we substitute back in \(u\) and then \(x\): First substitute back in \(u\): $$4\ln|u|+2\ln|u^2+2u+17|+C$$ Then substitute back in \(x\): $$4\ln|e^x|+2\ln|e^{2x}+2e^x+17|+C$$ Simplifying, we get: $$4x + 2\ln(e^{2x}+2e^x+17)+C$$ Thus, our solution is: $$\int \frac{68}{e^{2 x}+2 e^{x}+ 17} d x = 4x + 2\ln(e^{2x}+2e^x+17) + C$$

Key concepts

U-Substitution

When tackling integrals in calculus, u-substitution is a method that simplifies the process by making the expression easier to integrate. Imagine you have a complex expression where the presence of an exponent or a function within a function complicates the integral. U-substitution allows us to temporarily substitute part of the integral with a simpler variable, typically 'u'.

The trick lies in selecting the right part of the function to replace with 'u'. The goal is to make the new integral in terms of 'u' resemble a basic form that we can easily integrate. After integrating with respect to 'u', we then substitute back the original expression to get the final result. This technique is highly valuable when dealing with exponential functions, as in the provided exercise where we set u = e^x to simplify the integral.

Partial Fraction Decomposition

Another powerful technique in the integration arsenal is partial fraction decomposition. This method breaks down complex rational expressions into simpler fractions that are easier to integrate. It is particularly useful when dealing with ratios of polynomials.

The process involves expressing the rational function as a sum of simpler fractions, with the numerators typically being constants or linear functions, and the denominators being factors of the original denominator. Finding the constants requires setting up and solving a system of equations, which is an algebraic exercise. Once we have the decomposed form, which in our exercise is \(\frac{4}{u}-\frac{4u+8}{u^2+2u+17}\), each term can be integrated separately, often leading to the use of other integration techniques, including natural logarithms.

Integration Techniques

Calculus is rich with various integration techniques to navigate through the wide landscape of functions we might encounter. Beyond the straightforward power rule, techniques like integration by parts, trigonometric substitution, and the ones mentioned above — u-substitution and partial fraction decomposition — come into play.

Each technique has its own realm of effectiveness. For example, u-substitution is particularly good for handling composites of functions, while partial fraction decomposition is ideal for rational functions. The exercise given exemplifies the use of multiple techniques: u-substitution simplifies the integral, partial fractions break it down, and then the final integration may involve natural logarithms or inverse trigonometric functions, depending on the resulting terms.

Natural Logarithm

The natural logarithm, ln(x), is the inverse function of the exponential function e^x. In integration, the natural logarithm frequently appears when we integrate expressions in the form of 1/x or its derivatives. This is because the derivative of ln(x) is 1/x, making it a common result in integrals involving division by a variable.

In the context of the provided exercise, after partial fraction decomposition and u-substitution are performed, integrating a term like \(\frac{4}{u}\) naturally results in the natural logarithm, \(4\ln|u| + C\). The beauty of the natural logarithm lies in its simple derivative and integral, which makes it an integral part of solving calculus problems involving growth and decay, as well as many areas of integration.