Deriving an Energy Equation This is an exam problem that I've been doing most of the morning. I'm in college but unfortunately I live 100km away and I don't have anyone to help me and my exam is on Monday. So I'm doing past exam papers and just trying to get as much advice as possible. Thanks.  

A particle of mass $m=1$ moves along the x-axis under a conservative force $$F(x) = \dfrac{1}{x^2}- 1 $$
  (i) Starting with the equation of motion, derive the energy equation. 
(ii) Identify the potential energy $V(x)$. Plot $V(x)$ for $x>0$ clearly indicating the position of it's minimum. 
(iii) Find the total energy $E$ knowing that the particle has velocity $v=0$ when $x=3$. Find the maximal speed of the particle. 
(iv) Describe the type of motion. 

(i) Starting with the equation of motion $$F(x) = m\ddot{x}$$ Because $F(x)$ is dependent on displacement it makes sense to integrate with respect to $x$. We can multiply both sides by $\dot{x}dt$ as a handy way of integrating with respect to $x$. 
$$\begin{align}
   \int{F\dot{x}dt} & = \int{m\ddot{x}\dot{x}dt} \\
   \int{F\dfrac{dx}{dt}dt} & = \int{m\dfrac{d\dot{x}}{dt}\dot{x}dt} \\
   \int{Fdx}        & = \int{m\dot{x}d\dot{x}} \\
   \int{\dfrac{1}{x^2}-1 \, dx} & = \int{m\dot{x}d\dot{x}} \\
   \dfrac{1}{2}m\dot{x}^2 + \dfrac{1}{x}+x & = C
\end{align}
$$
Is multiplying both sides by $\dot{x}dt$ the best way to do this. The question actually gives a hint to rewrite $\dfrac{d^2x}{dt^2}$ using the chain rule. 
(ii)  $$V(x) = -\int{F(x)}$$
The potential energy  is the term $\Big(\dfrac{1}{x}+x\Big)$. 
Graph of $V(x)$ versus $x$. The minimum position is at  $x=1$.

(iii) The energy is defined by the energy equation that we found in part (i). 
$$ \dfrac{1}{2}m\dot{x}^2 + \dfrac{1}{x}+x  = E $$
Using our initial conditions we have 
$$\begin{align} 
E & =  \dfrac{1}{2}(1)(0)^2 + \dfrac{1}{3} + 3\\
\therefore \, \, \,  E & = \dfrac{10}{3} 
\end{align}$$
The maximum speed occurs when the kinetic energy is a maximum. Look at the energy equation it's easy to see that this occurs when $V(x)$ is at a minimum. We found on the graph that at $x=1$ the potential energy is a minimum. 
Hence, 
$$\begin{align}
\dfrac{1}{2}m\dot{x}^2 + \dfrac{1}{x}+x  & = E \\
(\dot{x}_m)^2 &= 2(\dfrac{10}{3}- 2) \\
\dot{x}_m & = \sqrt{2(\dfrac{10}{3}- 2)} \\
\dot{x}_m & = \dfrac{2\sqrt{6}}{3}
\end{align}$$
(iv) I found online that there is 4 types of motion. Rotation, Oscillatory, Linear, Irregular. This is the part I'm struggling with. The very first sentence of the problem says the particle moves along the x-axis. Does that mean that the motion is linear? Could I use that clue to justify my answer. Or is the description hidden in one of my answers. 
Thanks if you read through it. I appreciate all the help I can get. 
 A: *

*What yo have done is fine, although I would write "Multiplying by $\dot{x}$ and integrating wrt $t$" with similar working (it just looks more rigorous, even if it's the same process). I suspect the question wants you to use $$ \frac{d^2 x}{dt^2} = \frac{d\dot{x}}{dt} = \frac{dx}{dt} \frac{d\dot{x}}{dx} = \dot{x} \frac{d\dot{x}}{dx} = \frac{1}{2} \frac{d}{dx} \dot{x}^2 $$
and then integrate with respect to $x$. Both will give the same result, so just decide which you are more comfortable doing.

*You need a $dx$ on this integral. Should probably be an integration constant corresponding to where the potential energy is zero, since one can choose this arbitrarily (it goes away in the next part when considering the whole energy since then the integrals become definite, or if it's clearer, you're saying that the difference in the energy at different times is zero. Graph: the question may want the minimum labelled as $(1,2)$. Certainly there's no harm in adding more labels to a graph of this sort.

*Fine, as far as I can see. Worth saying explicitly "let $\dot{x}_m$ be the maximum value".

*I would say that the particle oscillates between the points where $x+1/x=3$ (which are probably worth finding explicitly since it's only a quadratic). This sort of descriptive question tends to require rather more than people initially expect. Also note that the graph is not just $a+bx^2$, so it isn't simple harmonic motion. I'm not sure that categorising the motion in the way you found is what the question actually wants: better to write a description with some key words sprinkled in it, the important point being that the particle moves backwards and forwards (explicitly, oscillates) between two specific points.
A: (i) It is better to multiply force by $dx$ since it makes sense that $Fdx = -dU$ for potential energy. In the left hand side you finally got this element but in the right hand side you have $\ddot{x}dx= \frac{d\dot{x}}{dt} \dot{x}dt$. 
(ii) In general there is an integration constant that is determined by the choice of zero point of potential.
(iii) is correct.
(iv) Your interpretation is not correct. Although the motion is one-dimensional, it can be oscillatory for example for 1D spring $m\ddot{x}=-kx$. 
You obtained $E = 10/3$ so the velocity is
$$ \dot{x} = \pm \sqrt{\frac{2}{m}(E - x- \frac{1}{x})}.$$
Since velocity cannot be a complex number, the expression under square root must be nonnegative, 
$$\dot{x}\in \mathbb{R} \quad \to \quad \frac{2}{m}(E - x- \frac{1}{x}) \ge 0 \quad \to \quad  x+ \frac{1}{x} \le \frac{10}{3}$$
Therefore the particle oscillates in the interval $[\frac{1}{3} , 3]$.
