Classical Point-Mass confined to a general quartic potential Consider a mass m confined to the x-axis with a potential energy of 
$$
U = kx^4
$$
Where
$$
k > 0
$$
at time $ t = 0$, when it is sitting at the origin, it is given a sudden kick to the right. (Positive $\hat{x}$ direction)   
a) Find the time for the mass to reach its maximum displacement $x_{max} = A$. Your answer will be given as an integral over $x$ in terms of $m$ , $A$, and $k$.  
b) Find the period $\tau$ of oscillations of amplitude $A$, and by making a suitable change of variables, show that it is inversely proportional to $A$. (Thus, the larger the amplitude the shorter the period!)  
c) The integral cannot be evaluated in terms of elementary functions. This is often the case, but for small oscillations about the minimum of any potential energy U(x), we may approximate U by the first three terms of its Taylor series in powers of $x$ (or if the minimum was located at $x = a$, in powers of $x-a$.
If
$$
U = \dfrac{x(x - 3)^2}{3}
$$
find the period of small oscillations about the minimum at $x = 3$. (Note that there is another equilibrium point at $x = 1$, but it is a maximum.)
I am stuck on part a) of this problem because of this I am expecting to need help throughout. I cannot figure out what the integral is supposed to be. What I have currently is the diffeq: (Obtained by taking the derivative of potential, and setting it equal to $ma$
$$
\ddot{x} = -4 \dfrac{k}{m} x^3
$$
Thank you in advance.
 A: I don't get part c) since this $U$ is unrelated to $kx^4$ but for a) and b):
(I'm doing the algebra on the fly so check for typos, but the method is sound.)
Remember this is a natural system with the kinetic energy $T=\frac{1}{2}m \dot{x}^2$ so the energy is conserved and 
$$
E=\frac{1}{2}m \dot{x}^2+kx^4\, .
$$
The energy $E$ can be evaluated at the turning point $x=x_{\small max}$, as at that point $\dot{x}=0$, so you now have
$$
\frac{2k}{m}(A^4-x^4)=\dot{x}^2
$$
or
\begin{align}
\frac{dx}{dt}&=\sqrt{\frac{2k}{m}(A^4-x^4)}\\
t_{max}&=\int_0^{x_{max}}dx \frac{1}{\sqrt{\frac{2k}{m}(A^4-x^4)}}\, ,
\end{align}
for the time taken to travel from $x=0$ to $x_{max}=A$.  This is one quarter of a full oscillation, so $\tau=4t_{max}$.
I'm pretty sure this is correct but Mathematica tells me the resulting integral is actually an elliptic $K$ function: to be clear:
$$
\displaystyle\int_0^A \frac{dx}{\sqrt{A^4-x^4}}= \frac{K(-1)}{A}
$$
which would produce $\tau\propto 1/A$.  However, according to this wiki article, Mathematica seems to have a variant of the usual notation.  To reach the form
$$
K(m)=\displaystyle\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m\sin^2(\theta)}}
$$
you need the substitution $x=A\sin(\theta)$.
I am confused with your part c): the fixed point for $U=kx^4$ is at $\dot{x}=0$ and $x=0$ since this is the minimum of the potential and where the momentum is $0$.  It seems this part is disconnected from parts a) and b), i.e. you need to redo a) and b) using $U= x(x-3)^2/3$ as your potential.
