Fact an easy limit but i don't get it Calculate:
$$\lim_{n\to\infty} \int_0^{n^2} \frac{\arctan x\cdot \ln x}{x(x^2+n^2)} \, dx.$$
I tried using $\ln x < x-1$ and $\arctan x < x$ but it's not working. Idk why.
 A: Using $\arctan(x)\leq\min(x,\pi/2)$ and $\log(x)\leq\sqrt{x}$ we have
$$\left|\int_{0}^{1}\frac{\arctan(x)\log(x)}{x(x^2+n^2)}\,dx\right|\leq \left|\int_{0}^{1}\frac{\log x}{n^2}\,dx\right|=\frac{1}{n^2},$$
$$0\leq \int_{1}^{n^2}\frac{\arctan(x)\log(x)}{x(x^2+n^2)}\,dx\leq\frac{\pi}{2}\int_{1}^{n^2}\frac{\sqrt{x}}{x(x^2+n^2)}\,dx\leq \pi\int_{0}^{+\infty}\frac{dz}{z^4+n^2}\leq\frac{C}{\sqrt{n}} $$
hence the limit is zero by squeezing.
A: The limit is $0$. This follows by DCT. Point-wise limit of the integrand  is $0$. In $(1,\infty)$ a dominating integrable function is $\frac \pi {2x^2}$ since $\ln x <x$. on $(0,1)$ a dominating integrable function is $C |\ln x|$ where $C$ is an upper  bound for $\left|\frac {\arctan x} x\right|$. [Note that $\frac {\arctan x} x \to 1$ as $x \to 0$].
Proof without using DCT: 
We can choose $N$ so large that $(\pi /2) \int_N^\infty \frac  1{x^2} \, dx <\varepsilon$. Since $|\arctan x |\leq \frac {\pi} 2$ and $x^2+n^2 \geq x^2$ it is clear that $\left|\int_N^\infty f_n(x) \, dx\right|<\varepsilon$ where $f_n$ is the given integrand. 
Now consider $\int_0^N f_n(x) \, dx$.  Here use the fact that $\frac {\arctan x } x$ is bounded, $\ln x \leq \ln N$ and $x^2+n^2 \geq n^2$. Can you complete the argument now?
A: $\arctan x \le \frac \pi 2\\
\ln x < x$
$\frac {(\arctan x)(\ln x)}{x}$ is bounded
Or  
$\frac {(\arctan x)(\ln x)}{x} < M$
$\int_1^{n^2}\frac {(\arctan x)(\ln x)}{x(x^2 + n^2)} \ dx < M \int_1^{n^2}\frac {1}{x^2 + n^2} \ dx$ 
$\int_1^{n^2}\frac {1}{x^2 + n^2} = \frac 1{n} \arctan \frac{x}{n}|_1^{n^2} < \frac {\pi}{2n}$
$0\le \lim_\limits{n\to \infty} \int_1^{n^2}\frac {(\arctan x)(\ln x)}{x(x^2 + n^2)} \le \lim_\limits{n\to \infty} \frac {M}{n} $
By the squeeze theorem, our limit equals $0.$
