At what values of the parameter a the antiderivative of the function has at most one local minimum. Function below. At what values of the parameter a the antiderivative of the function has at most one local minimum?
Function:

(x^4 -(3+a)x^3+(2+3a)*x^2-2ax)*exp(sin(x)/(x^2+3))

or you can look here - 1

But what to do next?

I understant that I need to consider f(x) as first derivative and then look when second derivative f'(x)>0 - which means that it's local minimum.

Thank you
 A: let $F(x)$ be the anti derivative of the given function. Then
$$F'(x) = (x^4 -(3+a)x^3+(2+3a)*x^2-2ax)*exp(sin(x)/(x^2+3)) $$
You can see clearly that $F'(x)=0$ at x = 0.
Now when is F''(x)>0 ? 
So,F"(x)>0 only when the function is "increasing" at $x$.
https://www.desmos.com/calculator/mtjvcgkkov
If you go on to this graph and play with $a$.
You will see how the graph behaves at $x=0$, first, the value of the function is $0$ for all values of 'a' at $x=0$. Secondly the function becomes increasing or in other words $F''(x)>0$ for $a<0$.
$$$$
The more conventional way is harder as 
$$F''(x) = e^{\frac{\sin (x)}{x^2+3}} \left(\frac{\left(x^2-3 x+2\right) x (x-a)
   \left(\left(x^2+3\right) \cos (x)-2 x \sin
   (x)\right)^2}{\left(x^2+3\right)^4}+\frac{2 \left(a \left(-3 x^2+6
   x-2\right)+x \left(4 x^2-9 x+4\right)\right) \left(\left(x^2+3\right) \cos
   (x)-2 x \sin (x)\right)}{\left(x^2+3\right)^2}+\frac{\left(x^2-3 x+2\right)
   x (a-x) \left(\left(x^4+15\right) \sin (x)+4 x \left(x^2+3\right) \cos
   (x)\right)}{\left(x^2+3\right)^3}-6 (a+3) x+6 a+12 x^2+4\right)$$ 
and 
$$F''(0) = -2a$$
Which is only positive for $a<0$
