$\int_0^\infty\frac{x\log(x)}{(1+x)^3}dx$ converges in making use of the sequence Show that the integral $\int_0^\infty\frac{x\log(x)}{(1+x)^3}dx$ converges in making use of the sequence 
$$f_n(x)=\begin{cases} 
   0 & \text{if } x\in[0,1/n) \\
   \frac{x\log(x)}{(1+x)^3}       & \text{if } x \in [1/n, \infty)
  \end{cases}$$
Could anyone give a strategy to solve this question? The first thing that comes to my mind is using the Lebesgue Dominated Convergence Theorem (LDCT). But, I couldn't see how to use it. Thanks!
 A: Note that the sequence $(f_n)$ point-wise converges to the function $f$ defined by $f(x):=\frac{x\log(x)}{(1+x)^3}$ for $x\in(0,\infty)$. If we find an integrable function $g$ on $(0,\infty)$ such that $|f_n(x)|\leq g(x)$ for all $n\in\Bbb N, x\in (0,\infty)$, then by the aid of LDCT,  we will have  $\int_0^\infty f(x)dx=\int_0^\infty\lim_{n\to\infty}f_n(x)dx=\lim_{n\to\infty}\int_0^\infty f_n(x)=\lim_{n\to\infty}\int_0^\infty |f_n(x)|\leq \int_0^\infty g(x)dx<\infty.$ 
Here we go:
First note that $\log(x)<\sqrt{x}$ for any $x\geq 1$ and we observe that, for any $n\in \Bbb N$, if $x\in (1/n,1]$ then 
\begin{align}
|f_n(x)|=\left|\frac{x\log(x)}{(1+x)^3}\right|=\frac{x|\log(x)|}{(1+x)^3}=\frac{x(-\log(x))}{(1+x)^3}=\frac{x\log(1/x)}{(1+x)^3} < \frac{x\sqrt{1/x}}{x^3}=x^{-5/2}.
\end{align} 
Also, if $x\in [1,\infty)$ then $|f_n(x)|=\frac{x\log(x)}{(1+x)^3}<\frac{x\sqrt{x}}{x^3}=x^{-3/2}$. So, for any $n\in \Bbb N$, we have 
$$|f_n(x)|<g_n(x):=\begin{cases} 
      \frac{1}{x^{5/2}} &,   x\in (\frac{1}{n},1] \\
      \frac{1}{x^{3/2}} &, x\in[1,\infty)
   \end{cases}$$
which implies that 
$$|f_n(x)|<g(x):=\lim_{n\to\infty}g_n(x)=\begin{cases} 
      \frac{1}{x^{5/2}} &,   x\in (0,1] \\
      \frac{1}{x^{3/2}} &, x\in[1,\infty)
   \end{cases}$$
