TITLE: Prove that the integral $\int^{10}_0 \frac{x}{x^3+16}dx$ is less than $\frac{5}{6}$
QUESTION [6 upvotes]: 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...

REPLY [11 votes]: 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 $\int^{\infty}_0 \frac{x}{x^3+16}\,dx$.