Let \(u = \ln x\) and \(dv = \frac{1}{x^4} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{3} x^{-3}\).
Using integration by parts, \(\int u \, dv = uv - \int v \, du\).
\(\int \frac{\ln x}{x^4} \, dx = -\frac{1}{3} x^{-3} \ln x + \frac{1}{3} \int x^{-3} \cdot \frac{1}{x} \, dx\).
\(= -\frac{1}{3} x^{-3} \ln x + \frac{1}{3} \int x^{-4} \, dx\).
\(= -\frac{1}{3} x^{-3} \ln x - \frac{1}{9} x^{-3}\).
Evaluate from 1 to \(a\):
\(\left[ -\frac{1}{3} x^{-3} \ln x - \frac{1}{9} x^{-3} \right]_{1}^{a}\).
\(= \left( -\frac{1}{3} a^{-3} \ln a - \frac{1}{9} a^{-3} \right) - \left( 0 - \frac{1}{9} \right)\).
\(= \frac{1}{9} - \frac{1}{3} a^{-3} \ln a - \frac{1}{9} a^{-3}\).
Since \(a > 1\), \(a^{-3} \ln a > 0\) and \(a^{-3} > 0\), thus \(\int_{1}^{a} \frac{\ln x}{x^4} \, dx < \frac{1}{9}\).
