Let \(I = \int\frac{1}{x(\log~x)^{2}}dx\)

Let \(u = \log~x\), then \(\frac{du}{dx} = \frac{1}{x}\), so \(du = \frac{1}{x}dx\)

Substituting into the integral, we get: \(I = \int\frac{1}{u^{2}}du = \int u^{-2}du\)

Using the power rule for integration, \(\int u^{n}du = \frac{u^{n+1}}{n+1} + c\), we have: \(I = \frac{u^{-2+1}}{-2+1} + c = \frac{u^{-1}}{-1} + c = -\frac{1}{u} + c\)

Substituting back \(u = \log~x\), we get: \(I = -\frac{1}{\log~x} + c\)