Evaluating a Lebesgue Integral

For fixed $x \in (0,1]$ the maximum of $\frac{n^{3/2}x}{1 + n^2x^2}$ is attained at $n = \frac{3^{1/2}}{x}$, providing an integrable dominating function:

$$\frac{n^{3/2}x}{1 + n^2x^2} \leqslant \frac{3^{3/4}}{4} \frac{1}{\sqrt{x}}$$

As pointed out by Botond, evaluating the integral readily leads to $$ \frac{1}{2\sqrt{n}} \, \ln \left( 1+n^2\right) \, $$ the limit of which is clearly $0$ as $n \to\infty$.

There is no function greater than $\frac{n^{1/2}}{2}$ for all $n$. So you will need a better upper bound. Your upper bound may depend on $x$, but not on $n$. Here, we will need an upper bound that ${}\to \infty$ as $x \to 0$.

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For example, it is enough to show $|f_n(x)| \le x^{-1/2}$ for all $n$ and all $x$, since $\int_0^1 x^{-1/2} dx$ converges.