Answer
$\dfrac{-9}{2}$
Work Step by Step
Let $f(x)=\int_{-8}^{1} \dfrac{1}{x^{1/3}} dx$
This can be re-written as: $f(x)=\int_{-8}^{0} \dfrac{1}{x^{(1/3)}} dx +\int_{0}^{1} \dfrac{1}{x^{(1/3)}} dx$
Now, $\lim\limits_{k \to 0^{-}}\int_{-8}^{0} \dfrac{1}{x^{(1/3)}} dx +\lim\limits_{k \to 0^{+}}\int_{0}^{1} \dfrac{1}{x^{(1/3)}} dx=\lim\limits_{k \to 0^{-}} [\dfrac{3}{2}(k)^{2/3}-6]+\lim\limits_{k \to 0^{+}} [(\dfrac{3}{2})-(\dfrac{3}{2})k^{(2/3)}]$
Thus, $\int_{-8}^{1} \dfrac{1}{x^{(1/3)}} dx=-6+\dfrac{3}{2}= \dfrac{-9}{2}$
