91Ó°ÊÓ

Integration by parts is a method used to integrate products of functions. The formula for integration by parts is similar to the product rule for derivatives and is expressed as: \[ \int u \, dv = uv - \int v \, du \].In practice, this technique involves choosing parts of the integral to assign to \(u\) and \(dv\), then computing \(du\) and \(v\) from these assignments. When selecting \(u\) and \(dv\), a common heuristic is to let \(u\) be the part that gets simpler when differentiated, and \(dv\) the part that can easily be integrated.In our exercise, the integral \( \int \frac{\ln(x)}{x} \, dx \) was solved using integration by parts. Here, \( u = \ln(x) \) (which simplifies to \( \frac{1}{x} \, dx \) when differentiated), and \( dv = \frac{1}{x} \, dx \) (which integrates to \( v = \ln(x) \)). Applying the integration by parts formula resolved the repeated evaluation to find a valid antiderivative.

Logarithmic properties are essential tools in calculus for simplifying expressions and making complex integrals more manageable. The properties used in the exercise include:

• Difference of logs: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \)
• Powers of logs: \( \ln(x^n) = n \cdot \ln(x) \)

These properties allowed the expression \( \ln(x^{11}) - \ln(x^{7}) \) to be simplified to \( \ln(x^4) \), which further broke down to \( 4 \ln(x) \). Simplifications like these are critical in transforming a complex-looking integral into a form that is easier to evaluate.

The antiderivative, also known as the indefinite integral, represents the process of reversing differentiation. It is a fundamental concept in calculus, providing a function whose derivative yields the original function being integrated.In the solution, finding the antiderivative of \( \int \frac{\ln(x)}{x} \, dx \) was made possible by using integration by parts, as discussed previously. The antiderivative was determined to be \( \frac{1}{2} \ln^2(x) + C \). Finding antiderivatives helps us to understand the area under curves and solve practical problems involving accumulation.

The constant of integration, denoted as \( C \), represents an arbitrary constant added to the function when finding an indefinite integral. Since differentiation of a constant is zero, integrating a function returns a family of functions differing by a constant. Therefore, \( C \) accounts for all possible vertical shifts of the antiderivative on a graph.In the exercise, once the antiderivative \( \frac{1}{2} \ln^2(x) \) was determined, \( C \) was included, resulting in the general solution \( \frac{1}{2} \ln^2(x) + C \). Incorporating \( C \) is crucial because it allows us to consider all potential solutions that could exist given the indefinite nature of the integral.