Answer
$$\frac{1}{{2{{\left( {\ln 4} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& \int_4^\infty {\frac{1}{{x{{\left( {\ln x} \right)}^3}}}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& {\text{Apply }}\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& \int_4^\infty {\frac{1}{{x{{\left( {\ln x} \right)}^3}}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_4^b {{{\left( {\ln x} \right)}^{ - 3}}\left( {\frac{1}{x}} \right)} dx \cr
& {\text{Integrate}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{{\left( {\ln x} \right)}^{ - 2}}}}{{ - 2}}} \right]_4^b = - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{{\left( {\ln x} \right)}^2}}}} \right]_4^b \cr
& {\text{Use the Fundamental Theorem of Calculus}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{{\left( {\ln b} \right)}^2}}} - \frac{1}{{{{\left( {\ln 4} \right)}^2}}}} \right] \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{{\left( {\ln b} \right)}^2}}} - \frac{1}{{{{\left( {\ln 4} \right)}^2}}}} \right] \cr
& {\text{Evaluate the limit}} \cr
& = - \frac{1}{2}\left[ {\frac{1}{{{{\left( {\ln \infty } \right)}^2}}} - \frac{1}{{{{\left( {\ln 4} \right)}^2}}}} \right] \cr
& {\text{Simplify}} \cr
& = - \frac{1}{2}\left( {\frac{1}{\infty } - \frac{1}{{{{\left( {\ln 4} \right)}^2}}}} \right) \cr
& = \frac{1}{{2{{\left( {\ln 4} \right)}^2}}} \cr} $$
