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?

As @Ryszard Szwarc mentioned in the comment, if $\frac{c_n}{n\ln n}\to\infty,\,\,\lim_{n→∞} \int _1^n\frac{\ln\:x}{c_n\:+\:x\:\ln x}dx=0$, and the most interesting is the case $c_n=n\ln n$.
So, we want to find $\lim_{n→∞} I(n)=\lim_{n→∞}\int _1^n\frac{\ln\:x}{n\ln n\:+\:x\:\ln x}dx$. Making the substitution $x=e^t$
$$I(n)=\int _1^n\frac{\ln\:x}{n\ln n\:+\:x\:\ln x}dx=\int_0^{\ln n}\frac{te^t}{n\ln n+e^t t}dt$$
Making another substitution $t=s\ln n$
$$I(n)=\frac{\ln n}{n}\int_0^1\frac{sn^s}{1+\frac{s}{n^{1-s}}}ds$$
As $\frac{s}{n^{1-s}}\leqslant1$, we can decompose the denominator into the series:
$$I(n)=\frac{\ln n}{n}\int_0^1n^ss\big(1-sn^{s-1}+s^2n^{2s-2}-s^3n^{3s-3}+-...\big)ds$$
Integrating by part, we get the terms with the different rate of convergence, for example:
$$\int_0^1sn^sds=\frac{1}{\ln^2n}\int_0^{\ln n}se^sds=\frac{1}{\ln^2n}\big(n\ln n-n\big)=\frac{n}{\ln n}-\frac{n}{\ln^2 n}$$
The same story happens to all the terms; only the first ones will give a non-zero limit at $n\to\infty$.
Therefore, keeping only these main terms
$$I(n)=\frac{\ln n}{n}\Big(\frac{1}{\ln^2n}\int_0^{\ln n}e^xxdx-\frac{1}{n\ln^3n}\frac{1}{2^3}\int_0^{2\ln n}e^xx^2dx+\frac{1}{n^2\ln^4n}\frac{1}{3^4}\int_0^{3\ln n}e^xx^3dx-+...\Big)$$
The pattern is clear. Integrating by part and keeping only first terms of every integral
$$I(n)=\frac{\ln n}{n}\Big(\frac{1}{\ln^2n}n\ln n-\frac{1}{n\ln^3n}\frac{1}{2^3}2^2n^2\ln^2n+\frac{1}{n^2\ln^4n}\frac{1}{3^4}3^3n^3\ln^3n-+...\Big)+O\Big(\frac{1}{\ln n}\Big)$$
$$\Rightarrow\,\,\boxed{\,\,\lim_{n\to\infty} I(n)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+-...=\ln 2=0.693147...\,\,}$$
The numeric evaluation of the limit confirms this conclusion:
$$\int _1^n\frac{\ln\:x}{n\ln n\:+\:x\:\ln x}dx\,\Big|_{n=10^{300}}=0.692143...$$
though the rate of convergence, of course, is monstrously low :)
It is not difficult to evaluate in the same way the second and third terms of the asymptotics. I got
$$\boxed{\,\,I(n)=\ln2-\frac{\ln2}{\ln n}+\Big(\ln2-\frac{\pi^2}{12}\Big)\frac{1}{\ln^2n}+O\Big(\frac{1}{\ln^3n}\Big)\,\,}$$
$\mathbf{PS}$
I checked numerically the revised asymptotics. It seems it works!
$$\int _1^n\frac{\ln\:x}{n\ln n\:+\:x\:\ln x}dx\,\Big|_{n=1000}=0.590148...$$
Asymptotics gives
$$I(n)\sim\ln2-\bigg(\frac{\ln2}{\ln n}+\Big(\ln2-\frac{\pi^2}{12}\Big)\frac{1}{\ln^2n}\bigg)\,\bigg|_{n=1000}=0.590093...$$
