Inspired by Hummus

let's generalize this

$\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ k ]{ x } } \right\} dx }$

substituting $x=\frac { 1 }{ { t }^{ k } }$

we get

$k\displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ t\} }{ { t }^{ k+1 } } dt } =k\displaystyle\sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { t-n }{ { t }^{ k+1 } } }dt }$

evaluating the integral and after simplifying a bit

$\frac { k }{ k-1 } \displaystyle\sum _{ n=1 }^{ \infty }{ \left( \frac { 1 }{ (n+1)^{ k-1 } } -\frac { 1 }{ { n }^{ k-1 } } \right) } +\displaystyle\sum _{ n=1 }^{ \infty }{ n\left( \frac { 1 }{ (n+1)^{ k } } -\frac { 1 }{ { n }^{ k } } \right) } =\\ \frac { k }{ k-1 } +\displaystyle\sum _{ n=1 }^{ \infty }{ \left( \frac { 1 }{ (n+1)^{ k-1 } } -\frac { 1 }{ { n }^{ k-1 } } \right) } -\displaystyle\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n+1)^{ k } } } =\frac { k }{ k-1 } -\zeta (k)$

then we substitute $k=6$

For simplicity, $f(x) = \frac{1}{x^{1/6}}$

$\displaystyle\int_{0}^{1}\{f(x)\}dx = \displaystyle\int_{0}^{1}f(x)dx - \lfloor f(x) \rfloor dx$

$\displaystyle\int_{0}^{1}f(x)dx$ obviously is $\frac{6}{5}$

$\displaystyle\int_{0}^{1}\lfloor f(x) \rfloor dx$ is evaluated as

$\displaystyle\sum_{i=1}^{\infty} \displaystyle\int_{i^{-6}}^{(i+1)^{-6}}idx = -\zeta(6) = \frac{-\pi^{6}}{945}$ as between each of these intervals $\lfloor f(x) \rfloor = i$

$\boxed{\displaystyle\frac{6}{5}-\frac{\pi^6}{945}}$