Proving the Indefinite Integral of 1/x in Calculus: A Detailed Guide

In this article, we will delve into the process of proving the indefinite integral of the function (frac{1}{x}) in calculus. Specifically, we will show that

[ int frac{1}{x} , text{d}x ln|x| ]

where the absolute value ensures the result is defined for all non-zero values of (x).

Introduction to the Problem

The goal is to demonstrate the integral of ( frac{1}{x} ) with respect to (x). Traditionally, this is written as

[ int frac{1}{x} , text{d}x ]

We will proceed with this proof in a step-by-step manner, ensuring clarity and precision in each step. For simplicity, we will omit the constant of integration, (C), which would be added in the final result.

Proof for (x > 0)

Consider the integral for positive values of (x)

[ int frac{1}{x} , text{d}x ]

For these values, we can make a substitution to simplify the integral.

Substitution: (x e^u)

Let (x e^u), where (u) is a new variable. Then, we have

[ text{d}x e^u , text{d}u ]

Substituting these into the integral, we obtain

[ int frac{1}{x} , text{d}x int frac{1}{e^u} cdot e^u , text{d}u int 1 , text{d}u u ]

Since (u ln |x|) and for (x > 0), (u ln x), we can write

[ int frac{1}{x} , text{d}x ln x ]

Proof for (x

Now, let's consider the case where (x) is negative, i.e., (x

Substitution: (x -e^u)

Let (x -e^u), where (u) is a new variable. Then, we have

[ text{d}x -e^u , text{d}u ]

Substituting these into the integral, we obtain

[ int frac{1}{x} , text{d}x int frac{1}{-e^u} cdot (-e^u) , text{d}u int 1 , text{d}u u ]

Since (u ln |x|) and for (x

[ int frac{1}{x} , text{d}x ln |x| ]

Note that the absolute value ensures the solution is valid for all non-zero (x).

Conclusion

In summary, we have shown that the indefinite integral of (frac{1}{x}) is given by

[ int frac{1}{x} , text{d}x ln |x| C ]

The constant (C) is added to represent the family of antiderivatives. This proof utilizes substitution to simplify the integral and then applies the properties of natural logarithms to derive the result.

Understanding this proof is crucial for many areas of calculus and its applications in differential equations, physics, and engineering.