Question

Resolve the apparent discrepancy between $$\begin{array}{c} \int \frac{d x}{x(x-1)(x+2)}=\frac{1}{6} \ln \frac{(x-1)^{2}|x+2|}{|x|^{3}}+C \text { and } \\\ \int \frac{d x}{x(x-1)(x+2)}=\frac{\ln |x-1|}{3}+\frac{\ln |x+2|}{6}-\frac{\ln |x|}{2}+C .\end{array}$$

Short answer

Answer: Yes, the expressions are equivalent when integrated. They represent the same function.

Step-by-step solution

Analyze the expressions

We need to prove that both expressions are equivalent$$\frac{1}{6} \ln \frac{(x-1)^{2}|x+2|}{|x|^{3}}+C = \frac{\ln |x-1|}{3}+\frac{\ln |x+2|}{6}-\frac{\ln |x|}{2}+C $$

Re-write using logarithm properties

We can use logarithm properties to rewrite each expression, considering the property$$\ln (a^n) = n \ln a$$ we have$$\frac{1}{6} \ln \big|\frac{(x-1)^{2}(x+2)}{x^{3}}\big|+C = \frac{\ln |x-1|}{3}+\frac{\ln |x+2|}{6}-\frac{\ln |x|}{2}+C $$It can also be written as$$\ln \big|\frac{(x-1)^{2}(x+2)}{x^{3}}\big|^{\frac{1}{6}}+C = \frac{\ln |x-1|}{3}+\frac{\ln |x+2|}{6}-\frac{\ln |x|}{2}+C $$

Apply logarithm rules

Using the logarithm property $$\ln a - \ln b = \ln \frac{a}{b}$$, rewrite the expression as $$\ln \big|\frac{(x-1)^{2}(x+2)^{\frac{1}{3}}}{x^{\frac{3}{2}}}\big|+C = \frac{\ln |x-1|}{3}+\frac{\ln |x+2|}{6}-\frac{\ln |x|}{2}+C $$ Now, apply the logarithm rule: $$a\ln x = \ln x^a$$ Applying this rule to the second equation: $$\ln \big|\frac{(x-1)^{2}(x+2)^{\frac{1}{3}}}{x^{\frac{3}{2}}}\big|+C = \ln \big|\frac{(x-1)^{\frac{1}{3}}(x+2)^{\frac{1}{6}}}{x^{\frac{1}{2}}}\big|+C $$

Compare the expressions

Now that we have simplified both expressions, we can easily compare them and see that they are equivalent.$$ \begin{array}{r} \ln \big|\frac{(x-1)^{2}(x+2)^{\frac{1}{3}}}{x^{\frac{3}{2}}}\big|+C \\ = \ln \big|\frac{(x-1)^{\frac{1}{3}}(x+2)^{\frac{1}{6}}}{x^{\frac{1}{2}}}\big|+C \end{array}$$ Since both expressions are equal when integrated, it's safe to say that they represent the same function. Therefore, there is no discrepancy between the two expressions, and they are indeed equivalent.

Key concepts

Logarithm Properties

Understanding logarithm properties is crucial for solving mathematical expressions that involve complex logarithmic terms. Logarithms are essentially the inverse of exponentiation, and their properties allow us to manipulate these expressions to simplify or solve them.

Key properties of logarithms that are often used in calculus include:

• **Product Property**: \( \ln(ab) = \ln a + \ln b \)
• **Quotient Property**: \( \ln \left(\frac{a}{b}\right) = \ln a - \ln b \)
• **Power Property**: \( \ln(a^n) = n \ln a \)

These properties allow us to break down or combine logarithmic expressions as seen in the given problem. The power property, in particular, is exploited to simplify expressions like \( \ln((x-1)^2) \) into \( 2 \ln |x-1| \), making them more manageable.
This simplification is fundamental when dealing with integrals involving logarithmic functions.

Integration

Integration is a core concept in calculus, playing a central role in computing areas under curves, among other applications. It's the reverse process of differentiation and is represented by the integral symbol \( \int \).

In the context of this problem, integration is used to solve for an expression in terms of a natural logarithm. The integral is taken over the function \( \frac{1}{x(x-1)(x+2)} \), and properties of integration are applied.

When integrating functions with logarithmic expressions, it is common to use methods such as partial fraction decomposition. This method breaks down complex rational expressions into simpler fractions that are easier to integrate.

Example techniques:

• **Substitution**: Simplifies an integral by replacing a complex part with a single variable.
• **Integration by parts**: Useful when the integrand is a product of two functions.

Understanding these methods is crucial for tackling integration problems that involve logarithms.

Definite Integrals

Definite integrals are a specific type of integral that compute the net area under a curve between two points. Unlike indefinite integrals, which result in a general expression plus a constant \( C \), definite integrals yield a specific numerical value.

In this problem, the solutions involve definite integrals as both expressions equate to an area under the given curve but with different appearances. While evaluating definite integrals, the properties of logarithms, such as combining or breaking down logs, are instrumental.

Steps in evaluating a definite integral:

• Determine the antiderivative.
• Apply the limits of integration.
• Compute the numerical result by subtracting the values at the upper and lower bounds of integration.

These operations make it possible to measure precise areas, which is a key utilization in engineering, physics, and economics. The correct manipulation and understanding of logarithm properties, as discussed, underlie successful evaluation of definite integrals in complex cases.