Answer
$${\text{False}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = 1? \cr
& {\text{The integrand is not defined for }}x = 0,{\text{ then}} \cr
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = \int_{ - 1}^0 {\frac{1}{{{x^3}}}} dx + \int_0^1 {\frac{1}{{{x^3}}}} dx \cr
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = \mathop {\lim }\limits_{b \to {0^ - }} \int_{ - 1}^b {\frac{1}{{{x^3}}}} dx + \mathop {\lim }\limits_{b \to {0^ + }} \int_a^1 {\frac{1}{{{x^3}}}} dx \cr
& {\text{Integrating}} \cr
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = \mathop {\lim }\limits_{b \to {0^ - }} \left[ { - \frac{1}{{2{x^2}}}} \right]_{ - 1}^b + \mathop {\lim }\limits_{b \to {0^ + }} \left[ { - \frac{1}{{2{x^2}}}} \right]_a^1 \cr
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = - \mathop {\lim }\limits_{b \to {0^ - }} \left[ {\frac{1}{{2{b^2}}} - \frac{1}{{2{{\left( { - 1} \right)}^2}}}} \right] - \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{1}{{2{a^2}}} - \frac{1}{{2{{\left( { - 1} \right)}^2}}}} \right] \cr
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = - \mathop {\lim }\limits_{b \to {0^ - }} \left[ {\frac{1}{{2{b^2}}} - \frac{1}{{2{{\left( { - 1} \right)}^2}}}} \right] - \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{1}{{2{a^2}}} - \frac{1}{{2{{\left( { - 1} \right)}^2}}}} \right] \cr
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = - \mathop {\lim }\limits_{b \to {0^ - }} \left[ {\frac{1}{{2{b^2}}} - \frac{1}{2}} \right] - \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{1}{{2{a^2}}} - \frac{1}{2}} \right] \cr
& {\text{Evaluate the limit}} \cr
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = - \left[ {\frac{1}{{2{{\left( {{0^ - }} \right)}^2}}} - \frac{1}{2}} \right] - \left[ {\frac{1}{{2{{\left( {{0^ + }} \right)}^2}}} - \frac{1}{2}} \right] \cr
& \int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = \infty \cr
& {\text{The integral diverges, therefore the statement }}\int_{ - 1}^1 {\frac{1}{{{x^3}}}} dx = 1{\text{ is}} \cr
& {\text{False}} \cr} $$
