Question: Which criterion do we use to show the convergence of $$\sum\_{n=1}^{\infty}\frac{n \log (n+1)}{1+n^3x^2}$$ ?

Do we maybe use that $\log (n+1)<n$ and so we get $\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^2}{1+n^3x^2}$ ?

Since $1+n^3x^2>n^3x^2$ we get $$\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^2}{n^3x^2}=\frac{1}{nx^2}$$ But this one diverges, right?

So do we have to find an other uppoer bound? Or do we use an other criterion?

Answer: 
You can use that $\log (n+1) \leqslant n^\alpha$ for any $\alpha \in (0,1)$ so you have estimation

$$\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^{1+\alpha}}{n^3x^2}=\frac{1}{n^{2-\alpha}x^2}$$

**Addition.**

As in comments arise question about continuity and differentiability, then let me say, that here key is domain with respect we are speaking. Function $f\_n(x)=\frac{n\ln (n+1)}{1+n^3x^2}$ is defined on $\mathbb{R}$ (for simplicity I considering $\ln$). Same with $f'\_n(x)=-\frac{2n^4x \ln (1+n)}{(1+n^3x^2)^2}$. Suppose we take any $x\_0>0$, then exists $x\_0 \gt a \gt 0$ such that $x\_0 \in [a,b]$ and using same technique as above gives us uniform convergence for $\sum\limits\_{n=1}^{\infty}f'\_n(x)$ on $[a,b]$, so differentiability of $f(x)=\sum\limits\_{n=1}^{\infty}f\_n(x)$ (so continuity) in $x\_0$.
As it was pointed out in comments, you can use a better bound for the logarithm:

$$
\dfrac{n \log(n+1)}{1+n^3 x^2} \leq \frac{n \sqrt{n}}{n^3 x^2}= \frac{1}{x^2} \frac{1}{n^{3/2}}
$$

which now yields a convergent series.