Answer
$\infty$
Work Step by Step
$\lim _{x\rightarrow \infty }\dfrac {\int ^{x}_{1}\ln \left( e^{4t-1}\right) dt}{x}=\dfrac {\dfrac {d}{dx}\left( \int ^{x}_{1}\ln \left( e^{4t-1}\right) dt\right) }{\dfrac {d}{dx}\left( x\right) }=\dfrac {\ln e^{4x-1}}{1}=\dfrac {4x-1}{1}=\infty $
