Problem 44 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the indefinite integral. $$\int \frac{e^{-u}-1}{e^{-u}+u} d u$$

Step-by-Step Solution

Answer

The indefinite integral of the given function is: \(\int \frac{e^{-u}-1}{e^{-u}+u} d u = \arctan(e^{-u}) - \frac{1}{2} \int \frac{e^{-2u}}{e^{-2u} + 1} d(e^{-u}) + C\)

Step 1: Simplify the given expression

Before integrating, let's try to simplify the given expression. We can do this by making a substitution. Let \(v = e^{-u}\), then \(-dv = e^{-u} du\). Now the integral becomes: $$\int \frac{v - 1}{v + \frac{1}{v}} (-dv)$$

Step 2: Distribute negative and simplify the expression

Now we will distribute the negative sign by multiplying the numerator by \(-1\), and also simplify the expression a bit more: $$\int \frac{1 - v}{v + \frac{1}{v}} dv$$ $$\int \frac{1 - v}{v^2 + 1} dv$$

Step 3: Separate the integrals

Now, we can break down the integral into two simpler integrals: $$\int \frac{1}{v^2 + 1} dv - \int \frac{v}{v^2 + 1} dv$$

Step 4: Integrate each expression separately

We recognize the first integral as the integral of the standard arctangent function: $$\int \frac{1}{v^2 + 1} dv = \arctan(v) + C_1$$ For the second integral, we can use a substitution (recall that \(v = e^{-u}\)). Let \(w = v^2\), then \(dw = 2v dv\). So, the integral becomes: $$\int \frac{v}{w + 1} \frac{1}{2}dw$$ $$\frac{1}{2} \int \frac{w^{1/2}}{w + 1} dw$$ Unfortunately, this integral cannot be simplified any further using elementary functions. So, we'll leave it as it is.

Step 5: Substitute back and combine the integrals

Now that we have found the individual integrals, let's substitute back and combine them: $$\int \frac{e^{-u}-1}{e^{-u}+u} d u = \arctan(e^{-u}) - \frac{1}{2} \int \frac{(e^{-u})^2}{(e^{-u})^2 + 1} d(e^{-u}) + C$$

Step 6: Simplify the expression

$$\int \frac{e^{-u}-1}{e^{-u}+u} d u = \arctan(e^{-u}) - \frac{1}{2} \int \frac{e^{-2u}}{e^{-2u} + 1} d(e^{-u}) + C$$ And this is our final answer.

Key Concepts

Integration Techniques

Integration is a central concept in calculus, useful for finding areas under curves, among other applications. Different integration techniques help solve various types of integrals.

• **Basic Integration Rule:** This is where you directly integrate simple functions using known formulas.
• **Substitution Method:** A technique used when the integral can be simplified by a suitable substitution.
• **Partial Fractions:** Useful when integrating rational functions.

Choosing the right technique often starts with analyzing the integral you have. If direct integration doesn't seem easy, substitution might be a good idea. Once the right technique is identified, solving integrals becomes more manageable. Consider which functions are inside the integral, and think about possible substitutions or transformations to simplify them.

Substitution Method

The substitution method is a powerful tool for transforming complicated integrals into simpler ones. By making an appropriate substitution, it becomes easier to evaluate the integral. This method involves:

• Choosing a suitable substitution variable (often a function inside the integral).
• Replacing the variable and differential in the integral with the new variable.
• Simplifying the integral and solving it.
• Substituting back the original variable to get the solution in terms of the initial variable.

For example, in the given exercise, let’s consider substituting \(v = e^{-u}\). This transforms the integral into a more manageable form. The original complex function simplifies into an expression with \(v\), making it easier to integrate. Substitution is about finding patterns and relationships that simplify the problem by changing the perspective.

Arctangent Function

The arctangent function, or \(\arctan(x)\), is the inverse of the tangent function. It helps in evaluating integrals involving the expression \(\frac{1}{x^2 + 1}\).
In calculus, recognizing forms like \(\int \frac{1}{x^2 + 1} \, dx\) lets us directly use the formula:\[ \int \frac{1}{x^2 + 1} \, dx = \arctan(x) + C \]This pattern shows up often, and knowing it can save a lot of time. Arctangent functions are key to many integration problems. In the exercise, recognizing this pattern allowed us to solve part of the problem quickly. Remember, these standard forms are essential to remembering and using efficiently.

Integral Simplification

Simplifying an integral can make it much easier to solve. The process often involves breaking down complex expressions into simpler pieces. **Simplification Tips:**

• Look for opportunities to factor expressions.
• Break integrals into smaller, manageable parts if possible.
• Use algebra to rearrange terms or combine like terms.

In our example exercise, splitting the integral into two simpler integrals made the problem more approachable:\[\int \frac{1}{v^2 + 1} dv - \int \frac{v}{v^2 + 1} dv\]This separation allowed us to tackle each part individually, using different known techniques for each. Reducing a complex integral into simpler forms is a smart way to find solutions effectively. Always aim to simplify before jumping into direct calculations.