uniformly convergent sequence of functions we have the next sequence of function:
$$ \forall x \ge 0, f_{n}(x)=\frac{nx} {e^x+n+x} $$
True or false - for all $$ b > a \ge 0 $$ we get: 
$$ \int_{a}^{b}(\lim_{n \to \infty}f_{n}(x))dx= \lim_{n \to \infty}\int_{a}^{b}f_{n}(x)dx $$
I think that the answer is -'true'. we have to prove that the sequence  of functions is uniformly convergent. 
 A: This is true.
See here for the proof that $(f_n)$ converges uniformly
(to the always vanishing function) on all compact interval of $[0,\infty)$, hence the limit commutes with the integral for all compact interval.
A: The answer is True, the limit does commute with the integral.
As Jean-Pierre Merx points out, you just need to show that 
$$g_n(x) = f_n(x) - x = - \dfrac{e^x+x^2}{e^x+n+x}$$ 
uniformly converges to 0. Take $|g_n(x)|$, and use some crude estimates.
$$|g_n(x)| = \dfrac{e^x+x^2}{e^x+n+x} < \dfrac{e^x+x^2}{n}$$
Now here is the important part: you are looking at a definite integral. You are integrating over some $[a,b]$, which means that by the Extreme Value Theorem, the numerator $e^x+x^2$ must achieve a maximum somewhere.
$$|g_n(x)| < \dfrac{M}{n}.$$
This $M$ will depend on $[a,b]$ (on $b$, to be more precise).  
As $n\to \infty$, $|g_n|\to 0$ uniformly on $[a,b]$. So the limit commutes with the integral.
Note: Uniform convergence implies that you can pass the limit through the integral. But the converse is not true -- you can have sequence of functions that do not necessarily converge uniformly, and still be able to pass the limit through the integral. 
