Problem 341 - Calculus Volume 1

Question

In the following exercises, find each indefinite integral by using appropriate substitutions. $$\int \frac{e^{\ln (1-t)}}{1-t} d t$$

Step-by-Step Solution

Answer

The integral is \\(-(1-t) + C\\).

Step 1: Identify the Inner Function

The integrand has a composition of functions where the exponential function involves a natural logarithm. Here, recognize the inner function in the exponent, which is \(u = \ln(1-t)\).

Step 2: Differentiate the Inner Function

Next, differentiate \(u = \ln(1-t)\) with respect to \(t\) to find \(du\). This gives \(du = \frac{-1}{1-t} dt\). We can rearrange this as \(dt = -(1-t) du\).

Step 3: Substitute and Simplify

Substitute \(u = \ln(1-t)\) and \(dt = -(1-t) du\) into the integral \(\int \frac{e^{\ln(1-t)}}{1-t} dt\). This gives: \[\int \frac{e^u}{1-t} \cdot -(1-t) du = -\int e^u du.\]Notice that \(1-t\) cancels out, simplifying the integral.

Step 4: Integrate with Respect to u

Now, integrate \(-\int e^u du\). The integral of \(e^u\) with respect to \(u\) is \(e^u\). Hence, \(-\int e^u du = -e^u + C\), where \(C\) is the constant of integration.

Step 5: Back Substitute the Original Variable

Since \(u = \ln(1-t)\), substitute back to get \(e^u = 1-t\). Thus, the solution is \(-(1-t) + C = -(1-t) + C\).

Key Concepts

Substitution Method

The substitution method is a powerful tool used in calculus to simplify complex integrals. When we encounter an integral that looks cumbersome or involves compositions of functions, substitution can make the process easier. By replacing part of the integrand with a simpler expression, we turn a complicated problem into a manageable one.
For the problem \( \int \frac{e^{\ln (1-t)}}{1-t} dt \), we use substitution to simplify. The key is identifying a part of the integral's expression that, when replaced, makes the integral easy to solve. In our case, we defined \( u = \ln(1-t) \). This substitution transforms the original variables into terms of \( u \), allowing for straightforward integration.
Here's how the substitution process works:

• Pick a substitution by identifying a complicated part of the integrand, like \( \ln(1-t) \) in this exercise.
• Differentiate the substitution variable to find \( du \) in terms of the original differential, \( dt \).
• Replace the original terms in the integral with the substitution terms.

By using this method, solving the integral becomes much simpler.

Integration Techniques

Integration techniques involve methods to solve integrals that may not be immediately obvious or straightforward. While some basic integrals can be found directly using standard forms, others require special strategies.
There are several integration techniques to handle different types of integrals:

• Substitution: Converts a complex integral into a simpler one, as seen in our problem.
• Integration by Parts: Useful for integrals that are products of two functions.
• Partial Fractions: Decomposes complex fractions into simpler parts, facilitating integration.

For the problem \( \int \frac{e^{\ln (1-t)}}{1-t} dt \), we applied substitution to transform the integrand into \( -\int e^u du \). Simplifying the integrand like this made it easy to apply a basic rule of integration, where the integral of \( e^u \) is \( e^u \).
This showcases the power of applying the right technique to solve integrals efficiently.

Exponential Functions

Exponential functions are a core concept in calculus, often encountered in various mathematical problems. These functions have the general form \( f(x) = a^{x} \), where \( a \) is a constant, usually \( e \), the base of natural logarithms.
In calculus, exponential functions have an interesting property: they are their own derivatives and integrals. This means the derivative of \( e^x \) is \( e^x \), and the integral of \( e^x \) is again \( e^x \) plus a constant of integration.
In the exercise \( \int \frac{e^{\ln (1-t)}}{1-t} dt \), the exponential function involves a natural logarithm, leading to \( e^{\ln(1-t)} \). Through substitution, the expression simplifies, letting us directly apply the rule that \( \int e^u du = e^u + C \). This property of exponential functions often simplifies calculations, making them easier to handle in various calculus problems.