Question

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

Short answer

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

Step-by-step solution

Identify the Substitution

Recognize that the expression can be simplified using substitution. Let \( u = 1 - t \). Then, the differential \( du = -dt \), or equivalently, \( dt = - du \).

Rewrite the Integral

Substitute \( u = 1 - t \) and \( dt = - du \) into the integral. The integral becomes \(-\int \frac{e^{\ln u}}{u} \, du\).

Simplify the Expression

Recognize that \( e^{\ln u} = u \) (since the exponential and the natural logarithm are inverse functions). Substitute this to simplify the integral to \(-\int \frac{u}{u} \, du = -\int 1 \, du\).

Integrate

The integral simplifies to \(-\int 1 \, du = -u + C\), where \( C \) is the constant of integration.

Substitute Back

Recall that \( u = 1 - t \), substitute \( u \) back to express the integral in terms of \( t \). This gives \(-(1 - t) + C\), which simplifies to \(-1 + t + C\).

Key concepts

Substitution Method

The substitution method is a crucial part of solving indefinite integrals, particularly when dealing with complex functions. This technique involves replacing a part of the integral with a new variable, typically to simplify the integration process.

• Identify a substitution where a segment of the integral can be replaced with a single variable, reducing complexity.
• Change the variable in the integral using the identified substitution and accompanying differential change.

In the exercise, the substitution method was employed by letting \( u = 1 - t \). This substitution simplifies the original integral by making the expression inside easier to work with.
Once the substitution is made, we also ensure to convert the differential \( dt \) into \( du \) format, which was correctly recognized as \( dt = -du \). This transformation aids in simplifying the integral into a solvable form.

Integration Techniques

Integration techniques are methods used to solve integrals, and they are essential in calculus. These techniques allow us to handle more challenging integrals through systematic steps.

• Direct Integration: Used when the integral is straightforward and matches a known standard integral form.
• Substitution: Useful for integrals where simplification is possible by changing variables.
• By Parts: Another technique handy for products of functions.

In this context, after substituting into \(-\int \frac{e^{\ln u}}{u} \, du\), the expression simplifies further using a property of exponentials and logarithms: \( e^{\ln u} = u \). This step is a direct use of known algebraic identities, a vital part of integrating functions that involve exponentials and logarithms.
After this simplification, the integral \(-\int 1 \, du\) is simply an integral of a constant, which is among the most basic of integrable functions. The result \(-u + C\) is straightforward, bringing the focus to substituting back, which is critical to expressing the answer in terms of the original variable.

Exponential Functions

Exponential functions are functions of the form \( e^x \), which are pivotal in calculus due to their unique properties, such as having the same derivative and integral. In our exercise, dealing with the expression \( e^{\ln (1-t)} \) involves an interesting use of logarithmic identities.

• Exponential and logarithmic functions are inverses. This means \( e^{\ln x} = x \) simplifying processes where these functions appear together.
• Their properties allow these functions to model continuous growth or decay, making them crucial in various applications, from population growth to radioactive decay.

For this problem, realizing that \( e^{\ln u} = u \) transforms a potentially complex situation into a much simpler one. This manipulation is not only mathematically sound but is often necessary in calculus to proceed with integration. Understanding how to apply the properties of exponential functions effectively allows one to master such calculus problems with assurance.