$\huge \begin{align} \displaystyle\int_{4}^{\,\infty} \frac{\ln\!\left(x\right)}{x} \d x &= \lim_{t\to \infty} \int_{4}^{t} \frac{\ln x}{x}\d x \\ \\\let u &= \ln(x) \\ \let \d u &= \frac{1}{x}\d \ x \\ &=\lim_{t\to \infty} \int_{4}^{t} {u \d u} \\&=\lim_{t\to \infty} \ba{ \frac{1}{2} u^2 }_{4}^{t} \\ \\&=\lim_{t\to \infty} \ba{ \frac{1}{2} u^2 }_{4}^{t} \\ \\&=\lim_{t\to \infty} \ba{ \frac{1}{2} \pa{\ln x}^2 }_{4}^{t} \\ \\&=\frac{1}{2} \lim_{t\to \infty} { {\ln 2x} } \big\rvert_{4}^{t} \\ \\&\text{Divergent.} \end{align} $