91Ó°ÊÓ

In calculus, a logarithmic expression often involves the natural logarithm, denoted as \( \ln \). When a logarithmic function is involved in an integral, it must sometimes be simplified using rules of logarithms. One such important property is \( \ln a^b = b \ln a \). This transformation helps reduce the complexity of the integral by expanding or simplifying the logarithm.

• This property is handy because it can turn products or powers within a logarithm into a more manageable form.
• In the given example, \( \ln(e^{2x-1}) \) leverages this rule.
• The exponent \( 2x - 1 \) is brought down in front of the \( \ln \) function, simplifying the expression to \( (2x - 1) \ln(e) \).

Since \( \ln(e) = 1 \), the expression \( (2x - 1) \ln(e) \) reduces further to just \( 2x - 1 \).
This simplification is crucial because it transforms a potentially complex logarithmic integral into a straightforward polynomial one.

The power rule for integration is a fundamental technique used when dealing with polynomials. It states that the integral of \( x^n \), where \( n eq -1 \), is \( \frac{x^{n+1}}{n+1} \) plus a constant.

• The power rule is applied to each term of the polynomial within the integral separately.
• In the problem above, the expression \( 2x - 1 \) is split into two terms: \( 2x \) and \( -1 \).
• For \( 2x \), treat it as \( 2x^1 \), applying the rule results in \( \int 2x \, dx = x^2 \).
• For \( -1 \), or \( -1x^0 \), it simplifies to \( \int -1 \, dx = -x \).

After applying the rule to each term, combine them to achieve the result \( x^2 - x \).
This method of integrating polynomials is straightforward and efficient, making it an essential tool in the integration toolkit.

Whenever you compute an indefinite integral, it's customary to add a constant of integration, denoted by \( C \). This constant is crucial because it represents the family of all antiderivatives derived from the integrated function.

• The indefinite integral gives a general solution, not a specific one.
• \( C \) adjusts for any vertical shifts in the antiderivative's graph.

In the example provided, after integrating \( 2x - 1 \), the result is \( x^2 - x \). Therefore, including the constant of integration, the complete solution is \( x^2 - x + C \).
This constant ensures the solution encompasses every possible antiderivative of a function, acknowledging that differentiation of any constant yields zero. Hence, the practical utility of \( C \) lies in solving real-world problems where initial conditions or specific solutions are necessary.