Answer
$\dfrac {255}{32}=7.96875$
Work Step by Step
$\int ^{4}_{1}\left( \dfrac {1}{x^{3}}+x\right) dx=\int ^{4}_{1}\left( x-3+x\right) dx=\left( \dfrac {1}{-3+1}x^{-3+1}+\dfrac {x^{2}}{2}\right) ]^{4}_{1}=(\dfrac {4^{-2}}{-2}+\dfrac {4^{2}}{2})-(\dfrac {1^{-2}}{-2}+\dfrac {1^{2}}{2})=\dfrac {255}{32}=7.96875$
