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...
 A: 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?
A: Let $x_n=\int_{n}^{2n}\frac{x+a}{x^3+2a}\,dx$.  Then, the first difference $x_{n+1}-x_n$ is given by
$$\begin{align}
x_{n+1}-x_n&=\int_{n+1}^{2n+2}\frac{x+a}{x^3+2a}\,dx-\int_{n}^{2n}\frac{x+a}{x^3+2a}\,dx\\\\
&=\int_{2n}^{2n+2}\frac{x+a}{x^3+2a}\,dx-\int_n^{n+1}\frac{x+a}{x^3+2a}\,dx\\\\
&=\int_n^{n+1}\left(\frac{4x+2a}{8x^3+2a}-\frac{x+a}{x^3+2a}\right)\,dx\\\\
&=-2\int_n^{n+1} \frac{2x^4+3ax^3-3ax-a^2}{(8x^3+2a)(x^3+2a)}\,dx
\end{align}$$
Now, if the sequence were decreasing, then $\int_n^{n+1} \frac{2x^4+3ax^3-3ax-a^2}{(8x^3+2a)(x^3+2a)}\,dx\ge 0$.  However, we see that the sequence is not decreasing for all $n\ge1$ in general.  
For example, take $a=100$.  Then, the integrand is negative for $n=1$, $n=2$, and $n=3$.  But, when $a=100$, the integrand is positive for $n\ge 4$ and hence the sequence $x_n$ is decreasing for $n\ge 4$.
In fact, with $a$ fixed, there exists a number $N$  large enough, so the $2x^4+3ax^3-3ax-a^2\ge 0$ whenever $x\ge N$.  So, for an arbitrary $a>0$, the best we can say is there exists a number $N$ such that $x_n$ is decreasing for $n\ge N$.
