Problem 68 - Calculus Single Variable

Question

Calculate the integrals. $$ \int \frac{\exp (x)}{1+\exp (x)} d x $$

Step-by-Step Solution

Answer

\( \ln|1 + e^x| + C \)

Step 1: Identify the Integral Type

The integral is in the form of a rational function where the numerator is the derivative of the denominator. This suggests we can use a substitution method to simplify the integration process.

Step 2: Choose a Substitution

Let us define a substitution to simplify the integrand. Set \( u = 1 + e^x \). Then the derivative \( du = e^x dx \). This helps us rewrite the integral in terms of \( u \).

Step 3: Rewrite the Integral

With the substitution \( u = 1 + e^x \) and \( du = e^x dx \), the integral becomes \( \int \frac{du}{u} \).

Step 4: Integrate the Simplified Integral

The integral \( \int \frac{1}{u} \, du \) is a standard logarithmic integral. Thus, the solution is \( \ln|u| + C \), where \( C \) is the constant of integration.

Step 5: Back-Substitute the Original Variable

Replace \( u \) with the original expression to return to the original variable: \( \ln|1 + e^x| + C \).

Step 6: Write Down the Final Solution

The solution to the integral \( \int \frac{e^x}{1+e^x} \, dx \) is \( \ln|1 + e^x| + C \), where \( C \) is the constant of integration.

Key Concepts

Substitution Method

The substitution method is a popular technique in calculus to simplify the integration of complex functions. It's like unraveling a puzzle by making things look simpler. Consider a situation where the function you're trying to integrate is not straightforward. That's when substitution comes in handy.
It's a method where we substitute a part of the integrand with a new variable. This makes the integral easier to solve.

• Choose an appropriate substitution: Pick a part of the integrand that can be replaced to make the integral easier. In our case, we used the substitution \( u = 1 + e^x \).
• Find the differential: Once you've chosen \( u \), differentiate it to find \( du \). For \( u = 1 + e^x \), the differential is \( du = e^x dx \).
• Rewrite the integral: Substitute \( u \) and \( du \) back into the integral. The integral becomes less complex and easier to evaluate.

This method turns a difficult integral into a simpler one, letting us solve it more easily.

Logarithmic Integration

Logarithmic integration comes into play typically when dealing with integrals of the form \( \int \frac{1}{u} \, du \). This type of integral is straightforward and results in a natural logarithm. It's a classic in calculus and serves as a foundation for solving more complex integrals.
In our exercise, once we simplified the integral with substitution, we arrived at a logarithmic form: \( \int \frac{du}{u} \). This was straightforward since the integral of \( \frac{1}{u} \) is \( \ln |u| + C \).

• The logarithmic integral results in a natural logarithm: \( \ln|u| \)
• The constant \( C \) is essential, as it accounts for any constant value that could be part of the original function.

Applying this technique efficiently wraps up the integration, bringing us to the neat logarithm-based solution.

Calculus Integrals

Integrals are a fundamental element of calculus, often described as the "inverse" of derivatives. They are crucial for calculating areas under curves and solving problems involving accumulation of quantities.
Understanding different types of integrals and methods for solving them is vital. For instance, a rational function like \( \frac{e^x}{1+e^x} \) might seem challenging at first glance. However, with techniques like substitution and logarithmic integration, these problems become manageable.
Some key integral types include:

• Definite Integrals: Calculate the accumulation of quantities over a specified range.
• Indefinite Integrals: Represent general form antiderivatives without specific limits.
• Improper Integrals: Deal with unbounded intervals or integrands.

The beauty of calculus integrals is revealed when complex problems are solved by understanding and applying correct techniques, like those used in our exercise.