Evaluate $\lim_{n\to\infty}\int_1^n\frac{\ln x}{c_n+x\ln x}\,dx$ Let $c_n$ be an unbounded sequence: $c_n\to\infty$ when $n\to\infty$
Find the value of the limit:
$$\lim_{n\to\infty}\int_1^n\frac{\ln x}{c_n+x\ln x}\,dx$$
What I managed to find was the fact that
$$\lim_{n\to\infty}\frac{\ln x}{c_n+x\ln x}=0$$
I was thinking  that the limit is going to be $0$, but I have to prove it still. Maybe it is not and I am wrong I do not know for sure.
I denoted
$$I(n)=\int_1^n\frac{\ln x}{c_n+x\ln x}\,dx$$
I was thinking of finding a recurrence formula for this integral, Though, I could not be able to make any significant progress.
What should I do?

It turns out that we have the following inequalities:
$$ \log\biggl(1 + \varliminf_{n\to\infty} \frac{n\log n}{c_n}\biggr)
\leq \varliminf_{n\to\infty} I(n)
\leq \varlimsup_{n\to\infty} I(n)
\leq \log\biggl(1 + \varlimsup_{n\to\infty} \frac{n\log n}{c_n}\biggr)
$$

Let $g : [0, \infty) \to [1, \infty)$ be the inverse of $x \mapsto x \log x$ for $x \geq 1$, that is, $g$ solves the functional equation
$$ g(t) \log g(t) = t $$
(In fact, we can write $g(t) = t/W(t)$ where $W(t)$ is the Lambert W-function. However, this observation is not needed here.) Then by substituting $t = x \log x$, or equivalently, $x = g(t)$, and noting that
$$ \frac{\mathrm{d}x}{x}
= \frac{g'(t) \, \mathrm{d}t}{g(t)}
= \frac{\mathrm{d}t}{g(t)(\log g(t) + 1)}
= \frac{\mathrm{d}t}{t + g(t)} $$
we have
\begin{align*}
I(n)
= \int_{1}^{n} \frac{x \log x}{c_n + x \log x} \, \frac{\mathrm{d}x}{x}
= \int_{0}^{n \log n} \frac{t}{c_n + t} \, \frac{\mathrm{d}t}{t + g(t)}
\end{align*}
Writing $\rho_n = (n \log n) / c_n$ and substituting $t = c_n u$,
\begin{align*}
I(n)
&= \int_{0}^{\rho_n} \frac{u}{1 + u} \cdot \frac{1}{u + g(c_n u)/c_n} \, \mathrm{d}u
\end{align*}
Now, let $ \displaystyle \alpha = \varliminf_{n\to\infty} \rho_n$ and $ \displaystyle 
\beta = \varlimsup_{n\to\infty} \rho_n$. Then
$$ \varlimsup_{n\to\infty} I(n)
\leq \varlimsup_{n\to\infty} \int_{0}^{\rho_n} \frac{u}{1 + u} \cdot \frac{1}{u} \, \mathrm{d}u
= \varlimsup_{n\to\infty} \log(1+\rho_n)
= \log(1+\beta) \tag{1} $$
On the other hand, together with the fact that $g(t) \ll t$ and the Fatou's lemma,
\begin{align*}
\varliminf_{n\to\infty} I(n)
&\geq \int_{0}^{\infty} \varliminf_{n\to\infty} \frac{u}{1 + u} \cdot \frac{1}{u + g(c_n u)/c_n} \mathbf{1}_{[0, \rho_n]}(u) \, \mathrm{d}u \\
&= \int_{0}^{\infty} \frac{u}{u+1} \cdot \frac{1}{u} \mathbf{1}_{[0, \alpha)}(u) \, \mathrm{d}u \\
&= \log(1+\alpha) \tag{2}
\end{align*}
