Lebesgue Dominated Convergence Theorem to find a limit Find the limits $\lim_{n \to \infty} \int_0^1 \frac{\ln(1+x)}{1+nx^2}$
This problem was given at the end of Lebesgue Integration Chapter. I assume I have to use use Lebesgue Convergence Theorem. In other words I have to find a function $g$ that dominates $f _n = \frac{\ln(1+x)}{1+nx^2)}$ then prove that $f_n$ converge to a function $f$. If I prove this then $\lim_{n \to \infty} \int_0^1 \frac{\ln(1+x)}{1+nx^2} = \int_0^1 f$
Any help on finding $g$ and $f$? Thanks
 A: If $n\in\mathbb N^*$ and $x\in[0,1]$
$$\left|\frac{\ln(1+x)}{1+nx^2}\right|\leq\frac{x}{1+nx^2}\leq \frac{1}{1+x^2}\in L^1(0,1)$$
Then you can use convergence dominated.
A: Convergence towards zero is a trivial consequence of the Cauchy-Schwarz inequality, since:
$$ \int_{0}^{1}\log^2(1+x)\,dx = 2\left(1-\log(2)\right)^2, $$
$$ \int_{0}^{1}\frac{dx}{(1+nx^2)^2} = \frac{1}{2}\left(\frac{1}{1+n}+\frac{\arctan\sqrt{n}}{\sqrt{n}}\right).$$
It is also true that:
$$ 0\leq \int_{0}^{1}\frac{\log(1+x)}{1+n x^2}\,dx \leq \int_{0}^{1}\frac{x}{1+n x^2}\,dx = \frac{\log(n+1)}{2n}.$$
A: We have the simple inequality 
$$1+nx^x\ge 1+x^2, n\ge 1 \tag 1$$
Thus, using $(1)$ reveals that
$$\left|\frac{\log (1+x)}{1+nx^2}\right|\le \frac{\log (1+x)}{1+x^2}$$
Since we have 
$$\int_0^1 \frac{\log (1+x)}{1+x^2}\,dx=\frac{\pi \log 2}{8}$$
then by the dominated convergence theorem
$$\begin{align}
\lim_{n\to \infty}\int_0^1\frac{\log (1+x)}{1+nx^2}\,dx&=\int_0^1\lim_{n\to \infty}\left(\frac{\log (1+x)}{1+nx^2}\right)\,dx\\\\
&=0
\end{align}$$
and we are done!
