Question

Evaluate each integral. $$\int \frac{e^{x}}{36-e^{2 x}} d x, x<\ln 6$$

Short answer

Question: Evaluate the integral $$\int \frac{e^{x}}{36 - e^{2x}} dx$$. Answer: The value of the integral is $$-\frac{1}{12}\ln|6-e^x| + \frac{1}{12}\ln|e^x| + C$$.

Step-by-step solution

Rewriting the denominator as a difference of squares

We want to take advantage of the structure of the denominator, so let's write the integral as: $$\int \frac{e^{x}}{(6 - e^x)(6 + e^x)} dx$$

Choosing a suitable substitution

Let's perform a substitution in the expression: Let \(u = 6 - e^x\). Then, \(du = -e^x dx\). Rearrange the formula of du to find \(dx\): $$dx = -\frac{du}{e^x}$$ Now substitute \(u\) and \(dx\) in the integral: $$-\int \frac{1}{u(u + 2e^x)} du$$ We need to get rid of the term \(e^x\) in the denominator. Use the original substitution \(u = 6 - e^x\), to solve for \(e^x\): $$e^x = 6 - u$$ Plug this expression of \(e^x\) into the integral: $$-\int \frac{1}{u(u + 2(6 - u))} du$$

Simplifying the integral

Now, simplify the integral: $$-\int \frac{1}{u(u + 12 - 2u)} du$$ $$-\int \frac{1}{u(12 - u)} du$$

Performing partial fraction decomposition

Perform partial fraction decomposition on the integral expression: $$\frac{1}{u(12-u)} = \frac{A}{u} + \frac{B}{12-u}$$ Solving for A and B, we get \(A = \frac{1}{12}\) and \(B = -\frac{1}{12}\). Replace these values in the integral: $$-\int \left(\frac{1}{12u} - \frac{1}{12(12 - u)}\right) du$$

Solving the integral

Now, we can evaluate the two separate integrals: $$-\frac{1}{12}\int \frac{1}{u} du + \frac{1}{12}\int \frac{1}{12 - u} du$$ Apply the natural logarithm rule on both integrals: $$-\frac{1}{12}\ln|u| + \frac{1}{12}\ln|12-u| + C$$

Re-substitute the original variable

Now we need to re-substitute the original variable \(x\) for \(u\). Recall that \(u = 6 - e^x\): $$-\frac{1}{12}\ln|6-e^x| + \frac{1}{12}\ln|6 - (6 - e^x)| + C$$ Finally, simplify the expression: $$-\frac{1}{12}\ln|6-e^x| + \frac{1}{12}\ln|e^x| + C$$ So the final answer is: $$\boxed{-\frac{1}{12}\ln|6-e^x| + \frac{1}{12}\ln|e^x| + C}$$

Key concepts

U-Substitution

In calculus, u-substitution is a method used to simplify the process of integration, especially when dealing with complex rational functions. It's particularly useful when you encounter functions within functions and can spot a piece of the integrand (the function you're integrating) whose derivative is present elsewhere within the integrand.

For example, with the given exercise, we chose a substitution that simplifies the denominator. By setting \(u = 6 - e^x\), we were able to find another term in the integral, \(e^x dx\), that related to its derivative, \(du\). The art of u-substitution lies in recognizing such relationships which, when substituted properly, transform the integral into a much simpler form. In complex cases, like the one we're looking at, it may even set the stage for additional techniques, such as partial fraction decomposition.

Partial Fraction Decomposition

The technique of partial fraction decomposition is applied to break down complex rational expressions into simpler fractions that can be integrated more easily. This method is especially useful when dealing with rational functions where the numerator is of lower degree than the denominator, as was the case in our exercise.

To execute this technique, we express the single complex fraction as a sum of simpler fractions with unknown coefficients, which we then solve for. In our example, we decomposed \(\frac{1}{u(12-u)}\) into \(\frac{A}{u} + \frac{B}{12-u}\) and found the values for \(A\) and \(B\) through a process of equation solving. This decomposition paved the way for straightforward integration of each term separately.

Indeterminate Forms

The term indeterminate forms refers to expressions in calculus that do not have a defined value without further evaluation. These often arise in the context of limits, but can also appear in integration, especially when integrating rational functions near points of discontinuity or around infinity.

In the exercise, an indeterminate form could potentially arise if \(e^x\) approaches a value such that both the numerator and the denominator tend to zero or infinity. However, note that the conditions of the exercise specify \(x < \ln 6\), meaning we never actually reach a point where the denominator is zero. Understanding indeterminate forms helps in avoiding incorrect assumptions about function behavior and is a fundamental part of evaluating integrals and limits correctly.

Natural Logarithm Integration

Integrating functions that result in the natural logarithm is a common outcome when working with the integration of reciprocal functions. In our exercise, after applying partial fraction decomposition, each term of the decomposed function represented the reciprocal of \(u\) or \(12 - u\), and both integrals follow the rule that the integral of \(1/x\) is \(\ln|x|\).

Since the natural logarithm function, \(\ln(x)\), is the inverse of exponentiation, we use this property to integrate functions like \(1/u\), leading to an expression that involves \(\ln|u|\). This step simplifies the integration process and provides a pathway to the solution, which, after re-substituting the original variable, completes the evaluation of the integral.