Prove that the integral $\int^{10}_0 \frac{x}{x^3+16}dx$ is less than $\frac{5}{6}$ Problem : 
Prove that the integral $\int^{10}_0 \frac{x}{x^3+16}dx$ is less than $\frac{5}{6}$
Getting no clue how to proceed as I am not getting any factor so that by manipulating I will be able to cancel the numerator or denominator : 
I the denominator could have been $x^3+32 = (x+4)(x^2-4x+16)$ could have been the factor and we will write numerator as x +4-4 but .... not getting the right clue here..Please suggest thanks...
 A: HINT: Recall $\int_a^bf(x)\mathrm{d}x\le(b-a)M$ where $f(x)\le M $ for $a\lt x \lt b$
(You can use calculus techniques to find $M$)
A: Some bounding of the function by easier to integrate functions gives stronger bounds than $\frac56$.  When $x$ is small, the upper bound $\dfrac{x}{16}$ is useful.  When $x$ is large, more useful is the upper bound $\dfrac{1}{x^2}$.  For each $a$ between $0$ and $10$, we have 
$$\int^{10}_0 \frac{x}{x^3+16}\,dx\leq\int_0^a\frac{x}{16}\,dx+\int_a^{10}\frac{1}{x^2}\,dx=\frac{a^2}{32}+\frac{1}{a}-\frac{1}{10}.$$  Because this is true for all $a$, it will hold when $a$ is chosen to make the last quantity as small as possible.  It has its minimum when $a=\sqrt[3]{16}$, giving a bound just under $\frac12$.  However, it would also be enough here to use the upper bound when $a=2$, which gives $$\int^{10}_0 \frac{x}{x^3+16}\,dx\leq\frac{4}{32}+\frac{1}{2}-\frac1{10}<\frac58<\frac56.$$
This would also work to give an upper bound of $\frac58$ (or a little smaller if we optimize) for $\int^{\infty}_0 \frac{x}{x^3+16}\,dx$.
A: Factor $ x^3 + 16  $
Note that if $x^3 + 16 = 0 $
Then:
$x^3 = -16$
And so therefore:
$x = 2*2^{1/3}$ times the three cube roots of unity
Lets call this number involving the 2's w for short
The roots of unity are 1,  (1 + i $3^{1/2}$)/2, and (1 - i $3^{1/2}$)/2 lets call these numbers 1, $o_1$ and $o_2$ respectively to keep it simple
Thus you now factor the x^3 + 16 as....
(x - 1w)(x - $o_1w$)(x - $o_2w$)
At this point a standard partial fraction decomposition approach will work.
Comment below if you want me to elaborate on that... Or you can do it yourself.
This of course is a brute force approach of actually solving the integral itself. I'm sure there are more elegant ways out there
