\(\int\left[(\log x-1) /(\log x)^{2}\right] d x=\ldots\) (a) \(x \log x\) (b) \(-x \log x\) (c) \([x /(\log x)]\) (d) \([(-\mathrm{x}) /(\log \mathrm{x})]\)

Let's choose the substitution function \(u = \log x\), which implies that \(x = e^u\).

Differentiate x with respect to u: \(\frac{dx}{du} = e^u\). Now, we can replace \(dx\) in our integral: \(dx = e^u \, du\).

Replace \(x\) and \(dx\) in the given integral using our substitution: \(\int \frac{(u - 1)}{u^2} e^u \, du\).

Now, let's integrate the new function with respect to u: \[\int \frac{(u - 1)}{u^2} e^u \, du = \int(e^u (\frac{1}{u^2}-\frac{1}{u})) du\] Now, let's split the integral into two parts: \[\int(e^u (\frac{1}{u^2}-\frac{1}{u})) du = \int \frac{e^u}{u^2} du - \int \frac{e^u}{u} du \] Now let's evaluate the two integrals separately. First integral: \(\int \frac{e^u}{u^2} du\) Using integration by parts, taking \(dv = \frac{e^u}{u^2} du\) and \(u = u\), we get, \(v = -\frac{e^u}{u}\) \(du = du\) Now let's substitute these in the expression: \[\int \frac{e^u}{u^2} du = -\frac{e^u}{u} - \int -\frac{e^u}{u} du = -\frac{e^u}{u} + \int \frac{e^u}{u} du \] Second integral: \(\int \frac{e^u}{u} du\) Let's use the result obtained from the first integral: \(\int \frac{e^u}{u} du = e^u/u + C\), where C is the constant of integration.

Now let's put it all together and substitute back \(x = e^u\) : \[-\frac{e^u}{u} + e^u/u + C = e^u\left(\frac{1}{u}\right) + C = x\left(\frac{1}{\log x}\right) + C\] So, \(\int\left[(\log x-1)/(\log x)^{2}\right] d x= x\left(\frac{1}{\log x}\right) + C\) Comparing with the choices: (a) \(x \log x\) (b) \(-x \log x\) (c) \([x /(\log x)]\) (d) \([(-\mathrm{x}) /(\log \mathrm{x})] \) Our answer matches with the choice (c).