how to use Lebesgue dominated convergence theorem to evaluate the limit of a definite integral?

How to solve out this Lebesgue integral?

$$\lim \limits_{n \to \infty} \int_0^1 (1+nx)(1+x^2)^{-n}dx$$

My attempt: I want to use Lebesgue dominated convergence theorem. If I can do so, then then I can switch the limit and integral. Then $\lim \limits_{n \to \infty} (1+nx)(1+x^2)^{-n} = \lim \limits_{n \to \infty} \frac{(1+nx)}{(1+x^2)^{n}}$.

Use L'Hôpital's rule $\frac{\infty}{\infty}$, we then get

$$\lim \limits_{n \to \infty} \frac{x}{(1+x^2)^n * \log(1+x^2)}$$

$$=\frac{x}{\log(1+x^2)} \lim \limits_{n \to \infty} \frac{1}{(1+x^2)^n}$$

$$= \frac{x}{\log(1+x^2)} *0=0$$

So I guess the final answer is 0? But I don't know how to find a dominating function $g(x)$.

Unfortunately, the minimal choice of the dominating function,

$$ g(x) = \sup_{n\geq 1} f_n(x) \qquad\text{for}\qquad f_n(x) = (1+nx)(1 + x^2)^{-n}, $$

is not integrable. This is because $g(x) \sim \frac{1}{ex}$ as $x \to 0^+$. The figure below demonstrates the graph of $f_n$ for $n = 1, \ldots, 100$ as well as the envelope of $f_n$'s.

To find the limit, apply the substitution $x = u/\sqrt{n}$ to get

$$ \int_{0}^{1} f_n(x) \, \mathrm{d}x = \int_{0}^{\sqrt{n}} \left(u + \frac{1}{\sqrt{n}} \right)\left(1 + \frac{u^2}{n}\right)^{-n} \, \mathrm{d}u. $$

Using the fact that $n \mapsto (1+u^2/n)^{-n}$ is decreasing for each $u$, it follows that

$$ \left(u + \frac{1}{\sqrt{n}} \right)\left(1 + \frac{u^2}{n}\right)^{-n} \leq \frac{u + 1/\sqrt{2}}{(1 + u^2/2)^2} $$

for all $n \geq 2$. Since this bound is integrable on $[0, \infty)$, by the dominated convergence theorem,

\begin{align*} \lim_{n\to\infty} \int_{0}^{1} f_n(x) \, \mathrm{d}x &= \lim_{n\to\infty} \int_{0}^{\sqrt{n}} \left(u + \frac{1}{\sqrt{n}} \right)\left(1 + \frac{u^2}{n}\right)^{-n} \, \mathrm{d}u \\ &= \int_{0}^{\infty} \lim_{n\to\infty} \left(u + \frac{1}{\sqrt{n}} \right)\left(1 + \frac{u^2}{n}\right)^{-n} \mathbf{1}_{[0,\sqrt{n}]}(u) \, \mathrm{d}u \\ &= \int_{0}^{\infty} u e^{-u^2} \, \mathrm{d}u \\ &= \left[ -\frac{1}{2}e^{-u^2} \right]_{0}^{\infty} = \frac{1}{2}. \end{align*}