Determine $ \int_{0}^{n}{\frac{\sin(x^n)}{x^n+x^{n+1}}dx} $ Show that $$\lim_{n\to\infty}{ \int_{0}^{n}{\dfrac{\sin(x^n)}{x^n+x^{n+1}}dx} }=\ln(2)$$
My approach: If the limits of integral is between 0 and 1, we have the integral approach to $\ln(2)$, but I don't concluded nothing when the limits of integrals are 0 and n. Thanks!  
 A: An elementary way, just to contrast with the nice real analysis argument: 
After the change of variables $u=x^n$ we have the integral
$$
I_n=\int_0^{n^n}\frac{u^{1/n-2}\sin u}{n(1+u^{1/n})}\,du
=\int_0^{1}\frac{u^{1/n-2}\sin u}{n(1+u^{1/n})}\,du
+\int_1^{n^n}\frac{u^{1/n-2}\sin u}{n(1+u^{1/n})}\,du=J_n+K_n.
$$
Now, for $J_n$, we use
$$
u-u^3/6\leq \sin u\leq u.
$$
Thus
$$
\frac1n\int_0^{1}\frac{u^{1/n-1}}{1+u^{1/n}}\,du-\frac{1}{6n}\int_0^{1}\frac{u^{1/n+1}}{1+u^{1/n}}\,du\leq J_n\leq \frac1n\int_0^{1}\frac{u^{1/n-1}}{1+u^{1/n}}\,du.
$$
But
$$
\frac1n\int_0^{1}\frac{u^{1/n-1}}{1+u^{1/n}}\,du=\bigl[\ln(1+u^{1/n})\bigr]_0^{1}= \ln 2,
$$
and
$$
0\leq \frac{1}{6n}\int_0^{1}\frac{u^{1/n+1}}{1+u^{1/n}}\,du
\leq \frac{1}{6n}
$$
and the right-hand side clearly tends to $0$ as $n\to+\infty$. It follows by the squeeze theorem for limits that $J_n\to\ln 2$ as $n\to+\infty$.
Next, we show that $K_n\to 0$ as $n\to+\infty$:
For $u>1$ one has
$$
\biggl|\frac{u^{1/n-2}\sin u}{n(1+u^{1/n})}\biggr|\leq\frac{1}{nu^2}.
$$
Thus
$$
0\leq |K_n|\leq \frac1n\int_0^{n^n}\frac{1}{u^2}\,du\leq \frac{1}{n}.
$$
It follows that $K_n\to 0$ as $n\to+\infty$.
We conclude that $I_n=J_n+K_n\to \ln 2$ as $n\to+\infty$.
Updated solution thanks to a mistake kindly pointed out by @felix-marin
A: Let $$g_n(x):=\frac{\sin\left(x^n\right)}{x^n}\frac 1{1+x}.$$
If $0\lt x\lt 1$, then $g_n(x)\to 1/(1+x)$. If $x\gt 1$, then $g_n(x)\to 0$. Moreover, for $n\geqslant 2$, 
$$\left|g_n(x)\right|\leqslant \frac 1{1+x}\mathbf 1_{(0,1]}(x)+\frac 1{x^2}\mathbf 1_{(1,+\infty)}(x).$$
