91Ó°ÊÓ

To evaluate the integral \( \int \frac{2}{v^2 - v} \ dv \), partial fraction decomposition is a useful technique to simplify the integrand. This method decomposes a complex rational function into simpler fractions. For our function, the denominator \( v^2 - v \) can be factored into \( v(v-1) \). This factorization allows us to express the original function as a sum of two simpler fractions:

• \( \frac{A}{v} \)
• \( \frac{B}{v-1} \)

To find the constants \( A \) and \( B \), equate \( \frac{2}{v(v-1)} = \frac{A}{v} + \frac{B}{v-1} \) and solve for \( A \) and \( B \). By multiplying through by the common denominator \( v(v-1) \), we derive a system of equations. Solving these gives us \( A = -2 \) and \( B = 2 \). Thus, our original integrand can be rewritten in terms of these partial fractions.

Once the integrand is decomposed into partial fractions, definite integration comes into play. The problem requires you to evaluate the integral from 2 to infinity. This essentially means finding the area under the curve defined by the integrand between these two points.After substituting the partial fractions back, we evaluate:

• \( \int_2^\infty \left( \frac{-2}{v} + \frac{2}{v-1} \right) dv \)

Breaking it down, integrate each part separately. Both resulting integrals involve natural logarithms, which are easier to manage once you properly apply integration limits. Remember, definite integration gives you a single numerical value, representing the total area under the curve between two specified points.

When approaching improper integrals that extend to infinity, we must consider the behavior by taking limits. In our case, \( v \) approaches infinity as it serves as the upper boundary of our integral:

• Evaluating the limit as \( v \rightarrow \infty \) affects the terms \( -2 \ln |v| \) and \( 2 \ln |v-1| \).

As \( v \) grows larger, both \( \ln |v| \) and \( \ln |v-1| \) tend to infinity, but their difference effectively cancels, resulting in 0 for the upper bound. This key observation makes the infinite bounds of integration manageable.

Integrating functions that yield natural logarithms is very common and requires understanding their behavior and properties. The natural logarithm \( \ln x \) involves a useful mathematical property: it converts multiplication into addition.For this problem, after decomposing into partial fractions, we integrate terms like \( \int \frac{-2}{v} \, dv = -2 \ln |v| \) and \( \int \frac{2}{v-1} \, dv = 2 \ln |v-1| \). These results rely on the integral formula \( \int \frac{1}{x} \, dx = \ln |x| \). Pay attention to the absolute values since logarithms are only defined for positive numbers.By understanding and applying these principles, you can evaluate the definite integral thoroughly—especially when using logarithmic properties to handle subtraction and addition during evaluation at limits.