Answer
$$
\int_{1}^{\infty} \frac{\ln x}{x^{4}} d x=\frac{1}{9}
$$
Work Step by Step
The given integral
$$
\int_{1}^{\infty} \frac{\ln x}{x^{4}} d x
$$
is an improper integral, hence
$$
\int_{1}^{\infty} \frac{\ln x}{x^{4}} d x =\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{\ln x}{x^{4}} d x
$$
Integrate by parts with
$$
\left[\begin{array}{ll}{u=\ln x,} & {d v=d x / x^{4}} \\ {d u=d x / x }, & { v=-1 / \left(3 x^{3}\right) }\end{array}\right]
$$
$$
\begin{aligned} \int_{1}^{\infty} \frac{\ln x}{x^{4}} d x &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{\ln x}{x^{4}} d x \\ &=\lim _{t \rightarrow \infty}\left[-\frac{\ln x}{3 x^{3}}\right]_{1}^{t}+\int_{1}^{t} \frac{1}{3 x^{4}} d x\\
&=\lim _{t \rightarrow \infty}\left(-\frac{\ln t}{3 t^{3}}+0+\left[\frac{-1}{9 x^{3}}\right]_{1}^{t}\right)
\quad\quad \left[ \text {let’s use L’Hospital’s Rule on the quotient.}\right] \\
&= \lim _{t \rightarrow \infty}\left(-\frac{1}{9 t^{3}}+\left[\frac{-1}{9 t^{3}}+\frac{1}{9}\right]\right) \\ &=0+0+\frac{1}{9}=\frac{1}{9} \end{aligned}
$$
