2. The function $f(x)=x+\frac{1}{(x+1)^{3}}+1(x>0)$. Then, the value of $x$ when the function reaches its minimum is

保留了源文本的换行和格式。
2. $\sqrt[4]{3}-1$.

Notice that,
$$
\begin{array}{l}
f(x)=x+\frac{1}{(x+1)^{3}}+1 \\
=\frac{1}{3}(x+1)+\frac{1}{3}(x+1)+\frac{1}{3}(x+1)+\frac{1}{(x+1)^{3}} \\
\geqslant 4 \sqrt[4]{\left(\frac{1}{3}\right)^{3}} .
\end{array}
$$

Equality holds if and only if $\frac{1}{3}(x+1)=\frac{1}{(x+1)^{3}}$, i.e., $x=$ $\sqrt[4]{3}-1$.