A non-linear integrable differential equation I'm trying to solve a question from a tutorial on mathematics for physicst which was never done due to the pandemic so I don't know the answer or a proper method of solving it. Nevertheless here is the question and my attempt at solving it. Feedback, suggestions on how to approach it and further readings recommendations would be hugely appreciated.
Let the equation of movement be:
$$m\ddot{x}(t) + V'(x(t))= 0\tag1$$
and,
$$E = \frac{m}{2}\left(\dot{x}(t)\right)^2 + V(x(t))\tag2$$
where $V(x)$ is a known derivable potential and $E$ is independent of $t$.

*

*By integration of the equation giving $\dot{x}$, Express the solution with the initial condition $x(t_0)=x_0$ in the form of t(x).

From the equation $(1)$
$$\dot{x}^2 = \frac{E-V}{m/2} \implies
\pm\int_\left(x_0\right)^x\sqrt{\frac{m/2}{E-V}}dx = t+Cste$$
Taking the positive root and from the initial condition we know $Cste=-t_0$
$$t(x)=\int_\left(x_0\right)^x\sqrt{\frac{m/2}{E-V}}dx+t_0$$
2. Let a increasing potential at infinity be:
$$V(x\rightarrow\infty)= \frac{-C}{x^\left(2a\right)}$$
where $C>0$ and $a>0$. We consider a particle of initial velocity $v_0>0$. Give the asymptotic behavior of $x(t)$ when $E>0$ and $E=0$.
I tried substituting the expression of $V(x)$ at infinity in the integral:
$$t(x)=\int_\left(x_0\right)^x\sqrt{\frac{m/2}{E+\frac{C}{x^\left(2a\right)}}}dx+t_0$$
I was trying to convert it in the form of $\arcsin(x)+c=\int\frac{1}{\sqrt{1-x^2}}dx$ by substitution but it has become apparent to me that it isn't possible perhaps I'm not allowed to directly substitute the expression of $V(x)$ at infinity.
I also think there is a way around this question without having to compute the integral but i can't seem to find one. Hope somebody can help me.
 A: I believe you answered the first question correctly however, the problem with the second question arise from the fact that you are trying to get the anti-derivative which in my opinion is very difficult.
Here's my approach:
Let us assume that x is near infinity then we have,
$$V(x\rightarrow\infty)= \frac{-C}{x^\left(2a\right)}$$
Let us replace this in the equation $(1)$ and integrate it:
$$\ddot{x}(t)=\frac{2aC}{m}x^\left(-2a-1\right)\\\implies\frac{x^\left(2a+3\right)}{2a(2a+2)(2a+3)}=\frac{C}{m}(t^2+C_1)$$
so we have:
$$x(t)=\frac{2aC}{m}(t^2+C_1)(2a+2)(2a+3)$$
also,
$$\dot{x}(t)=\frac{4aC(2a+2)(2a+3)}{m}t $$
let $D=a(2a+2)(2a+3)$,
$$\dot{x}(t)=\frac{4DaC}{m}t $$
replacing this in the equation $(2)$ since we want to introduce $E$ in the solution to study the asymptotic behaviour:
$$E = \frac{(4DaC)^2}{m^2}t^2 - \frac{C}{x^\left(2a\right)}\\
\implies x = \frac{1}{\sqrt[2a]{\frac{16C(Da)^2}{m^2}t^2-\frac{E}{C}}}$$
Here's a graph of $y = \frac{1}{\sqrt[2a]{x^2-Z}}$ (where $a$ and $Z$ are constants) to give you a better idea. Play with sliders to see the behavior of the function.
We can see from the graph that if $E=0$ a particle at position $x_1$ starts approaching $x=0$, which we can consider as the origin of potential, it takes an infinite amount of time for it to reach there(for most practical purposes we can consider as it being stopped). And, if $E>0$ same thing happens but the gap increases signifying that the particle is asymptotically stopped before it reaches the origin.
