Let $f(x)=\frac{e^x+1}{e^x-1}$ and $\int\limits_0^1 x^3 \frac{e^x+1}{e^x-1} d x=\alpha$. Then, $\int\limits_{-1}^1 t^3 f(t) d t$ is equal to

$2 \alpha$

Let $g(x)=x^3 f(x)$. Then,

$g(-x)=(-x)^3 f(-x)$

$\Rightarrow g(-x)=-x^3\left(\frac{e^{-x}+1}{e^{-x}-1}\right)=-x^3\left(\frac{e^x+1}{1-e^x}\right)=x^3\left(\frac{e^x+1}{e^x-1}\right)=g(x)$

$\Rightarrow g(x)$ is an even function.

Hence, $\int\limits_{-1}^1 t^3 f(t) d t=2 \int\limits_0^1 t^3 f(t) d t=2 \alpha$