$\sum_{n=0}^{+\infty} \frac{x^n}{1+nx} \stackrel{x \to 1^-}{\sim} -\ln(1-x)$

How to prove that $\sum_{n=0}^{+\infty} \frac{x^n}{1+nx} \stackrel{x \to 1^-}{\sim} -\ln(1-x)$ ?

First try : I tried to compare the sum with the integral $\int_0^{+\infty} \frac{x^t}{1+tx}\mathrm{d}t$. But I don't know how to calculate this integral.

Second try : $\forall x \in [0,1[$, $-\ln(1-x) = \sum_{n=1}^{+\infty} \frac{x^n}{n}$. So :\begin{align*}S(x) - \big(-\ln(1-x)\big) & = 1 + \sum_{n=1}^{+\infty} \left(\frac{x^n}{1+nx} - \frac{x^n}{n}\right) = 1 + \sum_{n=1}^{+\infty} \underbrace{\left(\frac{1}{1+nx} - \frac{1}{n}\right)}_{:=a_n(x)}x^n \end{align*}

My aim : try to show that $\sum_{n=1}^{+\infty}a_n(x)x^n \stackrel{x\to1^-}{=} \mathrm{O}(1)$.We have $$a_n(x) = \frac{n-1-nx}{n(1+nx)} \le \frac{n-nx}{n(1+nx)} = \frac{1-x}{1+nx} \le 1-x$$ thus $\sum_{n=1}^{+\infty}a_n(x)x^n \le x \le 1$. But it is not sufficient to proove $\sum_{n=1}^{+\infty}a_n(x)x^n = \mathrm{O}(1)$.