Đáp án:
$\begin{array}{l}
I = \int\limits_{ - 1}^0 {{{\left( {\dfrac{x}{{{x^4} + 1}}} \right)}^3}dx} \\
= \int\limits_{ - 1}^0 {\dfrac{{{x^3}}}{{{{\left( {{x^4} + 1} \right)}^3}}}dx} \\
= \dfrac{1}{3}\int\limits_{ - 1}^0 {\dfrac{{3{x^3}}}{{{{\left( {{x^4} + 1} \right)}^3}}}dx} \\
= \dfrac{1}{3}.\int\limits_{ - 1}^0 {\dfrac{1}{{{{\left( {{x^4} + 1} \right)}^3}}}d\left( {{x^4} + 1} \right)} \\
Đặt:{x^4} + 1 = a\\
\Rightarrow Khi:\left\{ \begin{array}{l}
x = - 1 \Rightarrow a = 2\\
x = 0 \Rightarrow a = 1
\end{array} \right.\\
\Rightarrow I = \dfrac{1}{3}\int\limits_2^1 {\dfrac{1}{{{a^3}}}da} \\
= \dfrac{1}{3}.\int\limits_2^1 {{a^{ - 3}}da} \\
= \left( {\dfrac{1}{3}.\dfrac{1}{{ - 2}}.{a^{ - 2}}} \right)_2^1\\
= \left( {\dfrac{{ - 1}}{{6{a^2}}}} \right)_2^1\\
= \left( {\dfrac{{ - 1}}{{{{6.1}^2}}} + \dfrac{1}{{{{6.2}^2}}}} \right)\\
= \dfrac{{ - 1}}{8}
\end{array}$
