14) Given the function $f(x)=\ln (x+1)+\frac{2}{x+1}+a x-2$ (where $a>0$).
(1) When $a=1$, find the minimum value of $f(x)$;
(2) If $x \in[0,2]$, $f(x) \geqslant 0$ always holds, find the range of the real number $a$.
$14 f(x)$ is defined on $(-1,+\infty)$.
(1) When $a=1$,
$$
\begin{array}{c}
f(x)=\ln (x+1)+\frac{2}{x+1}+x-2, \\
f^{\prime}(x)=\frac{1}{x+1}-\frac{2}{(x+1)^{2}}+1=\frac{x(x+3)}{(x+1)^{2}} .
\end{array}
$$

When $-10$, $f^{\prime}(x)>0$.
Therefore, $f(x)$ is a decreasing function on the interval $(-1,0]$, and an increasing function on the interval $[0,+\infty)$.
Thus, when $a=1$, the minimum value of $f(x)$ is $f(0)=0$.
(2) From (1), when $a=1$, $f(x) \geqslant 0$ always holds, i.e., $\ln (x+1)+\frac{2}{x+1}+$ $x-2 \geqslant 0$ always holds.
Therefore, when $a \geqslant 1$ and $x \in[0,2]$,
$$
\begin{aligned}
f(x) & =\ln (x+1)+\frac{2}{x+1}+x-2+(a-1) x \\
& \geqslant \ln (x+1)+\frac{2}{x+1}+x-2 \geqslant 0
\end{aligned}
$$

always holds.
Thus, $a \geqslant 1$ meets the requirement.
(2) When $0-1) .
\end{aligned}
$$

Since the discriminant of the equation $a x^{2}+(2 a+1) x+a-1=0$ is
$$
\Delta=(2 a+1)^{2}-4 a(a-1)=8 a+1>0,
$$

the equation $a x^{2}+(2 a+1) x+a-1=0$ has two distinct real roots, denoted as $x_{1}$ and $x_{2}$, with $x_{1}<x_{2}$.

From $x_{1} x_{2}=\frac{a-1}{a}<0$, we know that $x_{1}<0<x_{2}$. (The roots of $f^{\prime}(x)=0$ are one positive and one negative)

Therefore, when $0<x<x_{2}$, $f^{\prime}(x)<0$, and $f(x)$ is a decreasing function on the interval $\left[0, x_{2}\right]$.

If $0<x_{2}<2$, then $f\left(x_{2}\right)<f(0)=0$, which contradicts the condition that $f(x) \geqslant 0$ for $x \in[0,2]$.

If $x_{2} \geqslant 2$, then $f(2)<f(0)=0$, which also contradicts the condition that $f(x) \geqslant 0$ for $x \in[0,2]$.
Therefore, $0<a<1$ does not meet the requirement.
Combining (1) and (2), the range of values for $a$ is $[1,+\infty)$.