Evaluate $\lim_{a \to +\infty} \int_{a}^{a+1} \frac{x}{x+\ln x} \text{d}x$ I have to evaluate
$$\lim_{a \to +\infty} \int_a^{a+1} \frac{x}{x+\ln x} \, \text{d}x$$
My attempt: the limit exists because
$$\frac{\text{d}}{\text{d}x}\left(\frac{x}{x+\ln x} \right) = \frac{\ln x -1}{(x+\ln x)^2}$$
And since $a \to +\infty$ I can assume $a\geq e$, this makes
$$\frac{\ln x -1}{(x+\ln x)^2}\geq0$$
So the integrand is increasing, so for $a\geq e$
$$\frac{\text{d}}{\text{d}a}\left(\int_a^{a+1}\frac{x}{x+\ln x} 
 \, \text{d}x\right)=\frac{a+1}{a+1+\ln (a+1)}-\frac{a}{a+\ln a} \geq0$$
Hence, being the limit of an increasing function, it exists.
Now, since $a\geq e$, it is $\ln x \geq 0$ and $x \geq 0$, so
$$\int_a^{a+1} \frac{x}{x+\ln x} \, \text{d} x \leq \int_{a}^{a+1} \text{d}x=1$$
So we have the upper bound
$$\lim_{a \to +\infty} \int_a^{a+1} \frac{x}{x+\ln x} \, \text{d}x \leq 1$$
Now I would like to prove that the limit is $1$, but I haven't found a better lower bound of this one: using $\ln x\leq x-1$ for all $x>0$, it is
$$\lim_{a \to +\infty}  \int_a^{a+1} \frac{x}{x+\ln x} \, \text{d} x \geq \lim_{a \to +\infty}  \int_a^{a+1} \frac{x}{2x-1} \, \text{d} x = \frac{1}{2}$$
Actually I have some doubts about my attempt.
(1) Is it correct to assume $a \geq e$? I think it is possible because $a$ is going to $+\infty$, so it become greater than any fixed value.
So, in general, when I need to prove something about limits is it correct to assume things like this one? (for example, if $b \to -\infty$ can I assume that $b\leq 1$ or $b \leq -8449$?)
(2) Is my proof of the existence of the limit correct?
(3) How can I show that the limit is $1$? If it is $1$, maybe it is all wrong and the limit is not $1$. If it is, I would like to see a lower bound.
Thanks you all for your time.
 A: Your integral may be written
$$
1 - \int_a^{a + 1} {\frac{{\log x}}{{x + \log x}}dx} .
$$
Now note that
$$
\frac{{\log a}}{{a + \log a}} \le \int_a^{a + 1} {\frac{{\log x}}{{x + \log x}}dx}  \le \frac{{\log (a + 1)}}{{(a + 1) + \log (a + 1)}}.
$$
A: More general result: If $\lim_{x\to \infty}f(x) =L,$ then
$$\tag 1 \lim_{a\to \infty}\int_a^{a+1}f(x)\,dx=L. $$
In your problem $f(x)= x/(x+\log x),$ and $L=1$ (which you can prove using L'Hopital for instance).
To prove $(1),$ let $\epsilon>0.$ Then there exists $x_0$ such that $x>x_0$ implies $L-\epsilon<f(x)<L+\epsilon.$ Thus if $a>x_0,$ then
$$L-\epsilon=\int_a^{a+1}(L-\epsilon)\,dx < \int_a^{a+1}f(x)\,dx < \int_a^{a+1}(L+\epsilon)\,dx= L+\epsilon.$$
I.e., $|\int_a^{a+1}f(x)\,dx-L|<\epsilon,$ proving $(1).$
A: You're overthinking things. Much simpler is to note that
$${a\over a+\ln a}\le{x\over x+\ln x}\le1$$
if $x\le a\ge e$, since $u/(x+\ln u)=1/(1+(\ln u)/u)$ and $f(u)=(\ln u)/u$ is decreasing for $u\ge e$.  It follows that
$${a\over a+\ln a}\le\int_a^{a+1}{x\over x+\ln x}dx\le1$$
and now the Squeeze Theorem does the rest, since
$${a\over a+\ln a}={1\over1+(\ln a)/a}\to{1\over1+0}=1$$
Remark: You don't really need to know that $(\ln u)/u$ is decreasing, you really only need to know that its limit is $0$ as $u\to\infty$. That is, you can get the same result for the integral of $x/(x+\sin x\ln x)$, for example. The argument, however, is somewhat subtler: the simple inequality $a/(a+\sin a\ln a)\le x/(x+\sin x\ln x)$ is no longer always true, so you have to modify it to an inequality that is always true.
