Determine the integral of the function from 3 to infinity: $\int_{3}^{\infty} \frac{6}{v^{2}-v} d v$

Step 1 ：First, simplify the integrand by factoring the denominator: \(v^{2}-v = v(v-1)\). So, the integrand becomes: \(\frac{6}{v(v-1)}\).

Step 2 ：Perform partial fraction decomposition to further simplify the integrand: \(\frac{6}{v(v-1)} = \frac{A}{v} + \frac{B}{v-1}\).

Step 3 ：Multiplying through by the common denominator gives: \(6 = Av - A + Bv\).

Step 4 ：Setting \(v = 0\) gives \(A = -6\). Setting \(v = 1\) gives \(B = 6\). So, the integrand becomes: \(-6\frac{1}{v} + 6\frac{1}{v-1}\).

Step 5 ：Now, integrate term by term from 3 to infinity: \(\int_{3}^{\infty} -6\frac{1}{v} dv + \int_{3}^{\infty} 6\frac{1}{v-1} dv\).

Step 6 ：The integral of \(1/v\) is \(\ln|v|\), and the integral of \(1/(v-1)\) is \(\ln|v-1|\). So, we have: \(-6[\ln|v|]_{3}^{\infty} + 6[\ln|v-1|]_{3}^{\infty}\).

Step 7 ：As \(v\) approaches infinity, \(\ln|v|\) also approaches infinity. However, as \(v\) approaches infinity, \(\ln|v-1|\) also approaches infinity. So, both terms are indeterminate in the form of \(\infty - \infty\).

Step 8 ：To resolve this, use L'Hopital's rule, which states that the limit of a quotient of two functions that both approach infinity or both approach zero is equal to the limit of the quotients of their derivatives.

Step 9 ：Applying L'Hopital's rule to the first term gives: \(-6\lim_{v\to\infty} \frac{1}{v} = 0\).

Step 10 ：And applying L'Hopital's rule to the second term gives: \(6\lim_{v\to\infty} \frac{1}{v-1} = 6\).

Step 11 ：So, the integral from 3 to infinity of the given function is \(6 - 0 = 6\).

Step 12 ：Therefore, the final answer is \(\boxed{6}\).