$\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x=$
$\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x=$
$a^{2}$
0
$a$
$2 \int_{0}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x$
Solution
Let
$\begin{aligned}
f(x) &=x^{2}\left[\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right]=x^{2}\left[\frac{e^{x^{3}}-\frac{1}{e^{x^{3}}}}{e^{x^{3}}+\frac{1}{e^{x^{3}}}}\right]=x^{2}\left[\frac{\left(e^{x^{3}}\right)^{2}-1}{\left(e^{x^{3}}\right)^{2}+1}\right] \\
f(-x) &=(-x)^{2}\left[\frac{e^{-x^{3}}-e^{x^{3}}}{e^{-x^{3}}+e^{x^{3}}}\right]
\end{aligned}$
$=x^{2}\left[\frac{\frac{1}{e^{x^{3}}}-e^{x^{3}}}{\frac{1}{e^{x^{3}}}+e^{x^{3}}}\right]=x^{2}\left[\frac{1-\left(e^{x^{3}}\right)^{2}}{1+\left(e^{x^{3}}\right)^{2}}\right]=-x^{2}\left[\frac{\left(e^{x^{3}}\right)^{2}-1}{1+\left(e^{x^{3}}\right) 2}\right]=-f(x)$
Thus $f(-x)=-f(x) \Rightarrow$ Given function is an odd function.
$\therefore \quad I=0$