Integration by Parts Problem: Help in understanding why a part of it equals 0 $$4I= \int_0^{\infty} \frac{4x^3 +\sin(3x)-3\sin x}{x^5} \ \mathrm{d}x $$ $$=\frac{-1}{4} \underbrace{\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}}_{=0} +\frac{1}{4} \int_0^{\infty} \frac{12x^2 -3\cos x +3\cos(3x)}{x^4} \ \mathrm{d}x $$
Now, I was told that the brackets gives zero, but not why it is so. Could somebody please help explain this to me? Many thanks!

Let's evaluate
$${\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}} =$$
$$\lim_{x\to \infty}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )-\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$
take a look on $$\lim_{x\to \infty}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$
 $sin(3x)$ and  $3sin(x)$ are a bounded functions and the deg($4x^3$)< deg($x^4$).
Then you'll find that the value of this limit, when x goes to infinity, is cero.
What about, 
$$\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$
It's obvious that the functions $f(x)=4x^3+\sin(3x)- 3 \sin x$  and $g(x)={x^4}$ cab be derived four times so if we apply L'Hopital theorem 4 times we'll find that
$$\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )=\lim_{x\to 0}(\frac{\ 81sin(3x)- 3 \ sinx}{24} )=0$$
In conclusion $${\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}} =0-0=0$$
