Prove that $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$ is decreasing Let $\ a>0$ and the sequence $(x_{n})_{n>=0}$ defined by $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$. Prove the sequence is monotonically decreasing and $0<x_n<\frac{4+3a}{8}$, for any $n>0$. I've made little progress towards proving that $x_{n}<x_{n-1}$. First, if a function takes the form of the fraction , $f(x)=\frac{x+a}{x^{3}+2a}$, then it's monotonically decreasing. Also, $x_n$ compared to $x_{n-1}$ boils down to comparing $2\frac{x+2a}{x^{3}+16a}$ with $\frac{x+a}{x^{3}+2a}$. From here I have really no idea how to continue. These simple results are the work of some ruminations and I am unable to bring something out of the blue to complete the proof...

First, let's note that 
$$ \int_n^{2n} \frac{x+a}{x^3+2a}dx \leq \int_n^{2n} \frac{x+a}{x^3}dx.$$
Evaluating the integral on the right, we have 
$$ \int_n^{2n} \frac{x+a}{x^3}dx = \int_n^{2n} \frac{1}{x^2}dx + a\int_n^{2n} \frac{1}{x^3}dx.$$
These are easy integrals to evaluate. This gives us that 
$$ \int_n^{2n} \frac{x+a}{x^3}dx = \frac{3a + 4n}{8n^2} \leq \frac{3a + 4}{8}$$
for $n \geq 1$. At $n = 0$, we have that the integral is 
$$\int_0^0 \frac{x+a}{x^3+2a}dx = 0,$$
and so denoting 
$$ I_n = \int_n^{2n} \frac{x+a}{x^3+2a}dx,$$
we get
$$ 0 \leq I_n \leq \frac{3a + 4}{8}$$
for all $a > 0$. Now, we would like to establish
$$ I_{n+1} \leq I_n.$$
Can you take it from here?
