91Ó°ÊÓ

To check if the integral converges or diverges, we need to look at its integrand and the limits of integration. It is given as follows: $$\int_{1}^{\infty} \frac{d v}{v(v+1)}$$ Observe that the integrand is a rational function of \(v\), and the limits of integration are from \(1\) to \(\infty\). We can proceed to the next step of breaking the integrand into simpler fractions, to determine convergence.

We rewrite the integrand using partial fractions decomposition as follows: $$\frac{1}{v(v+1)}=\frac{A}{v}+\frac{B}{v+1}$$ Multiplying both sides by the common denominator \(v(v+1)\), we get: $$1=A(v+1)+B(v)$$ Solving simultaneously for \(A\) and \(B\) by setting \(v=0\) and \(v=-1\) gives us \(A=1\) and \(B=-1\). Thus, the integrand can be written as: $$\frac{1}{v(v+1)}=\frac{1}{v}-\frac{1}{v+1}$$

Now, we can rewrite the original integral as: $$\int_{1}^{\infty} \frac{d v}{v(v+1)}=\int_{1}^{\infty}\left(\frac{1}{v}-\frac{1}{v+1}\right)d v$$ Separate and evaluate each integral: $$\left[\log\left|v\right|-\log\left|v+1\right|\right]_{1}^{\infty}$$ Note that \(-\infty < \log x < \infty\), so the integral converges as the limits of integration lead to finite values. Evaluating the difference of these logarithms at the limits gives: $$\lim_{t\to\infty}\left[\log\left|\frac{t}{t+1}\right|-\log\left|\frac{1}{2}\right|\right]$$ Use properties of logarithms and simplify: $$\lim_{t\to\infty}\left[\log\frac{t}{t+1}-\log\left(\frac{1}{2}\right)\right]=\log1-\log\frac{1}{2}=\boxed{\log2}$$ So, the value of the given integral is \(\log2\).