Problem 38 - Applied Calculus

Question

Find the integrals in problems. Check your answers by differentiation. $$ \int \frac{e^{t}}{e^{t}+1} d t $$

Step-by-Step Solution

Answer

The integral is \( \ln|e^t + 1| + C \).

Step 1: Substitution

First, use the substitution method to simplify the integral. Set \( u = e^t + 1 \), which implies \( du = e^t \, dt \). This means that \( dt = \frac{du}{e^t} \). Therefore, substituting into the integral we have the following: \[ \int \frac{e^t}{e^t + 1} \, dt = \int \frac{1}{u} \, du. \]

Step 2: Integral of 1/u

The integral \( \int \frac{1}{u} \, du \) is a standard form, which results in a logarithmic function. So, the integration gives: \[ \int \frac{1}{u} \, du = \ln|u| + C, \] where \( C \) is the constant of integration.

Step 3: Back-Substitution

Substitute back the value of \( u \) that was initially assumed as \( u = e^t + 1 \). Therefore, the integral becomes: \[ \ln|e^t + 1| + C. \]

Step 4: Differentiation to Check the Answer

Differentiate the obtained result \( \ln|e^t + 1| + C \) to verify the solution. Differentiating \( \ln|e^t + 1| \) with respect to \( t \) gives: \[ \frac{d}{dt} \ln|e^t + 1| = \frac{1}{e^t + 1} \cdot e^t = \frac{e^t}{e^t + 1}, \] which is the original integrand, confirming our correct integration.

Key Concepts

Substitution Method

The substitution method transforms a complex integral into a simpler one, which makes integration more manageable.
In our exercise, we made the substitution by setting a new variable,\( u = e^t + 1\). This substituted variable \( u \) represents a more straightforward expression.
This step is crucial because it allows us to rewrite the integral in terms of \( u \) instead of the original variable \( t \). In this example, we found that the derivative \( du \) is equal to \( e^t \, dt \). We use these relationships to change variables in our integral:

• Determine \( u \) from the expression.
• Find \( du \) by differentiating \( u \) with respect to \( t \).
• Substitute \( u \) and \( du \) into the integral, replacing all occurrences of \( t \) and expressions involving \( dt \).

The substitution simplifies the integral to a better-known form, in this case, \( \int \frac{1}{u} \, du \). After integration is performed, the final step is back-substitution, where we convert back to the original variable \( t \), using our initial substitution \( u = e^t + 1 \).

Logarithmic Integration

Logarithmic integration occurs when you have an integral of the form \( \int \frac{1}{x} \, dx \).
This kind of integral results in a logarithm function, which is useful for simplifying expressions that might initially seem daunting. In our exercise, the substitution led us to the integral:
\( \int \frac{1}{u} \, du \).
Recognizing this as a standard form, the integral evaluates to:

• \( \ln|u| \) plus the constant of integration \( C \).

Logarithmic integration is straightforward once you spot the \( \frac{1}{x} \) pattern. After finding \( \ln|u| + C \), back-substitute \( u \) for the original variable expression, giving \( \ln|e^t + 1| + C \).
Such transformations help in resolving complicated integrals through standard techniques.

Differentiation

Differentiation serves as a reliable method to check the accuracy of an integral.
After finding the integral, you differentiate the resulting expression to see if it returns to the original integrand.
In our case, the result \( \ln|e^t + 1| + C \) was differentiated. Differential calculus rules applied here involve:

• The chain rule, which helps in differentiating composed functions \( \ln(f(t)) \).
• Simplifying the derived expression back to \( \frac{e^t}{e^t + 1} \).

This final result matches the original function that was inside the integral, confirming that our integration was performed correctly. Differentiation gives us confidence in our solutions, verifying that the problem has been addressed appropriately.