Problem 38 - Applied Calculus

Question

Find each indefinite integral by the substitution method or state that it cannot be found by our substitution formulas. $$ \int \frac{1}{1-x} d x $$

Step-by-Step Solution

Answer

The indefinite integral is \(-\ln|1-x| + C\).

Step 1: Identify potential substitution

In this integral, let's consider a substitution to simplify the integrand. We notice the expression \(1-x\) in the denominator. We can use the substitution \( u = 1 - x \).

Step 2: Compute the differential

Find the differential \( du \) in terms of \( dx \). If \( u = 1 - x \), then \( du = - dx \). Therefore, \( dx = -du \).

Step 3: Rewrite the integral with substitution

Substitute \( u = 1 - x \) and \( dx = -du \) into the integral. We have:\[ \int \frac{1}{u} (-du) = - \int \frac{1}{u} du. \]

Step 4: Integrate the substituted function

Integrate \(- \int \frac{1}{u} du\):\[ - \ln |u| + C, \] where \(C\) is the constant of integration.

Step 5: Reverse substitution

Substitute back \( u = 1-x \) to express the integral in terms of \( x \):\[ - \ln |1-x| + C. \]

Key Concepts

Substitution Method

The substitution method serves as a fundamental tool for solving indefinite integrals, providing a way to simplify the integrand. The central idea of this method is to replace a complicated or cumbersome expression with a simpler one through a substitution. Think of it as a way of changing variables to make the integration process easier.

Here's how you can think of substitution method:

• Choose a substitution: Look for a part of the integrand that, if replaced, will simplify the problem. Often, it's a function inside another function or an expression that can be easily differentiated.
• Find the differential: Once you've chosen a substitution, find the differential of your new variable in terms of the old variable.
• Rewrite the integral: Substitute both the variable and the differential into the integral.

After carrying out these steps, you \( \textbf{integrate} \) with your new variable. Then, you backtrack and reverse the substitution to return your variable to the original one used in the integration problem.

In practical terms, substitution can be likened to a mini ‘change of scenery’ for the function, shifting perspectives to simplify the process.

Integration by Substitution

Integration by substitution, often viewed as a special case of the chain rule from differentiation, is where the substitution method is practically applied. The process is often like undoing the derivative of a composition of functions, called the chain rule, in reverse.

Steps to perform integration by substitution:

• Identify a composite function: Typically a function nested within another, like \( f(g(x)) \).
• Substitute and differentiate: Let \( u = g(x) \) and determine \( du = g'(x) dx \). Effectively, you're replacing \( dx \) with \( du/g'(x) \).
• Change the variable: Rewrite the integral in terms of \( u \) and carry out the integration.

In our example, we used the expression \( u = 1-x \). This substitution transformed the integral into a simpler form that could easily be integrated: \( -\int \frac{1}{u} du \). Once integrated, it's essential to revert to the original variable to obtain the final solution.

By focusing on simplification and clever manipulation of variables, integration by substitution targets what might otherwise seem complex integrals and transforms them into something manageable and solvable.

Inverse Functions

Inverse functions are functions that reverse the effect of the original function. If you have a function \( f(x) \) and its inverse \( f^{-1}(x) \), applying \( f^{-1} \) to \( f(x) \) brings you back to the original input, \( x \). In the context of integration, understanding inverse functions helps in breaking down integrals, especially those involving logarithmic and exponential functions.

These are valuable in the substitution method:

• Recognizing inverse relationships allows you to swap in substitutes that relate to the inverted counterparts.
• Depth in understanding leads you to rely on known integrals, like the logarithmic integral \( \int \frac{1}{x} dx = \ln |x| + C \).

In the solution, the integral \( \int \frac{1}{u} du \) correlates to the inverse of the exponential function—logarithms. The result \(-\ln |1-x| + C\) is an instance where logarithms, coupled with integration, reveal inverse behavior.

By knowing and recognizing these inverse functions, challenging integrals can often be simplified or recognized as patterns, ultimately connecting to fundamental calculus operations.