(i) To find \(\int \frac{\ln x}{x^3} \, dx\), use integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{x^3} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{2x^2}\).
Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:
\(\int \frac{\ln x}{x^3} \, dx = \left( \ln x \right) \left( -\frac{1}{2x^2} \right) - \int \left( -\frac{1}{2x^2} \right) \left( \frac{1}{x} \right) \, dx\)
\(= -\frac{\ln x}{2x^2} + \int \frac{1}{2x^3} \, dx\)
\(= -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C\)
(ii) To evaluate \(\int_1^2 \frac{\ln x}{x^3} \, dx\), substitute the limits into the expression obtained:
\(\left[ -\frac{\ln x}{2x^2} - \frac{1}{4x^2} \right]_1^2\)
\(= \left( -\frac{\ln 2}{8} - \frac{1}{16} \right) - \left( -\frac{0}{2} - \frac{1}{4} \right)\)
\(= -\frac{\ln 2}{8} - \frac{1}{16} + \frac{1}{4}\)
\(= -\frac{\ln 2}{8} + \frac{1}{8}\)
\(= \frac{1}{8} - \frac{\ln 2}{8}\)
\(= \frac{1}{8}(1 - \ln 2)\)
Using \(\ln 2 = \frac{1}{2} \ln 4\), we have:
\(= \frac{1}{8}(1 - \frac{1}{2} \ln 4)\)
\(= \frac{1}{16}(2 - \ln 4)\)
\(= \frac{1}{16}(3 - \ln 4)\)
