$\lim_{n \rightarrow \infty}\int_0^1f_n(x)$ For $n=1, 2,...,$ let $f_n(x)=\frac{2nx^{n-1}}{x+1}, x\in [0, 1]$. Then $\lim_{n \rightarrow \infty}\int_0^1f_n(x)$
Function is unbounded at $1$, How do I solve?
 A: \begin{align*}
\int_{0}^{1}f_{n}(x)dx&=\int_{0}^{1}\dfrac{2nx^{n-1}}{x+1}dx\\
&=\dfrac{2}{x+1}\cdot x^{n}\bigg|_{x=0}^{x=1}-\int_{0}^{1}-\dfrac{2}{(x+1)^{2}}\cdot x^{n}dx\\
&=1+2\int_{0}^{1}\dfrac{x^{n}}{(x+1)^{2}}dx,
\end{align*}
now
\begin{align*}
\dfrac{x^{n}}{(x+1)^{2}}\leq\dfrac{1}{(x+1)^{2}},~~~~x\in[0,1],
\end{align*}
and
\begin{align*}
\int_{0}^{1}\dfrac{1}{(x+1)^{2}}=-\dfrac{1}{x+1}\bigg|_{x=0}^{x=1}=\dfrac{1}{2}<\infty,
\end{align*}
and 
\begin{align*}
\dfrac{x^{n}}{(x+1)^{2}}\rightarrow 0,~~~~x\in[0,1),
\end{align*}
so by Lebesgue Dominated Convergence Theorem,
\begin{align*}
\int_{0}^{1}f_{n}(x)dx\rightarrow 1.
\end{align*}
Another way:
\begin{align*}
\int_{0}^{1}\dfrac{x^{n}}{(x+1)^{2}}dx\leq\int_{0}^{1}x^{n}dx=\dfrac{1}{n+1}\rightarrow 0.
\end{align*}
A: Since $\int_0^1nx^{n-1}dx=1$, we have
$$\begin{align}
0
&\le\left|\int_0^1f_n(x)dx-1 \right|\\
&=\left|\int_0^1{2nx^{n-1}\over1+x}dx-\int_0^1nx^{n-1}dx \right|\\
&=\left|\int_0^1\left({2nx^{n-1}\over1+x}-{n(1+x)x^{n-1}\over1+x} \right)dx \right|\\
&=\left|\int_0^1{n(x^{n-1}-x^n)\over1+x}dx\right|\\
&=\int_0^1{n(x^{n-1}-x^n)\over1+x}dx\quad\text{(since the integrand is clearly nonnegative)}\\
&\le\int_0^1n(x^{n-1}-x^n)dx\\
&=1-{n\over n+1}\\
&={1\over n+1}\to0
\end{align}$$
So by the Squeeze Theorem,
$$\lim_{n\to\infty}\int_0^1f_n(x)dx=1$$
