Answer
$${\text{The integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{{x^3}}}{{{x^4} - 1}}dx} \cr
& {\text{The integrand is not defined for }}x = 1,{\text{ then}} \cr
& \int_0^1 {\frac{{{x^3}}}{{{x^4} - 1}}dx} = \mathop {\lim }\limits_{a \to {1^ - }} \int_0^a {\frac{{{x^3}}}{{{x^4} - 1}}dx} \cr
& {\text{Integrating}} \cr
& = \frac{1}{4}\mathop {\lim }\limits_{a \to {1^ - }} \left[ {\ln \left| {{x^4} - 1} \right|} \right]_0^a \cr
& = \frac{1}{4}\mathop {\lim }\limits_{a \to {1^ - }} \left[ {\ln \left| {{a^4} - 1} \right| - \ln \left| {{0^4} - 1} \right|} \right] \cr
& = \frac{1}{4}\mathop {\lim }\limits_{a \to {1^ - }} \left[ {\ln \left| {{a^4} - 1} \right|} \right] \cr
& {\text{Evaluating the limit}} \cr
& = - \infty \cr
& {\text{The integral diverges}} \cr} $$
