<s>[INST] Find the intervals in which the function decreasing. $f(x)= \log{(1+x)}-{2x\over 1+x^2}$ $f(x)= \log{(1+x)}-{2x\over 1+x^2}$  find the intervals in which the function decreasing.
My solution: $$f'(x)={1\over 1+x}- {2(1-x^2) \over (1+x^2)^2} \\ ={x^4+2x^3+4x^2-2x-1 \over (1+x)(1+x^2)^2}$$
Since we need to find the interval for decreasing hence $f'(x)<0$.
Therefore, $${x^4+2x^3+4x^2-2x-1<0}$$
Now how to proceed to find the interval. Thanks for help.
 [/INST] The first derivative must be less than zero for a function to be decreasing.
$f'(x)={x^4+2x^3+4x^2-2x-1 \over (1+x)(1+x^2)^2}$,
The system of inequalities must be resolved by:
$x^4+2x^3+4x^2-2x-1 <0$,
$x+1>0$,
for the existence of the logarithm in the real field.
In addition, it must be $x≠-1$.
The solution to the problem is as follows:
$x_{1}<x<x_{2}$,
$x_{1}=-\frac{\sqrt{6}\sqrt{3\sqrt{R}-S}}{6}+\frac{\sqrt{3T}}{6}-\frac{1}{2}$,
$x_{2}=+\frac{\sqrt{6}\sqrt{3\sqrt{R}-S}}{6}+\frac{\sqrt{3T}}{6}-\frac{1}{2}$;
the values of $R$, $S$, and $T$ are:
$R=4A^{1/3}+4B^{1/3}+4a^{1/3}+4b^{1/3}+1$,
$S=6C^{1/3}+6c^{1/3}+5$,
$T=12D^{1/3}+12d^{1/3}-5$;
and again
$A=\frac{4375}{5832}-\frac{125\sqrt{129}}{1944}$,
$a=\frac{4375}{5832}+\frac{125\sqrt{129}}{1944}$,
$B=\frac{2386}{729}-\frac{70\sqrt{129}}{243}$,
$b=\frac{2386}{729}+\frac{70\sqrt{129}}{243}$,
$C=\frac{35}{216}-\frac{\sqrt{129}}{72}$,
$c=\frac{35}{216}+\frac{\sqrt{129}}{72}$,
$D=C$,
$d=c$.
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