What are some tips/techniques that might help me solve this (brutal) differential equation? I've been working on a certain physics problem involving differential equation for two years. I've made some progress on it recently, but I've come across another roadblock, namely an integral that I have no idea how to compute. What are some tips or techniques that might help me evaluate it? I've been having trouble with it because I've nev taken a diff-eq class, although I'm also aware that this is a difficult problem in itself (non-linear second order equation). Please note: I do NOT want the solution the the problem, just the tools to solve it myself.
Here's the problem:

There is a fixed massive body at the origin and another object on the x axis with some initial velocity in the x direction. Find an equation that describes the position of the orbiting object with respect to time.

Here is my work so far. Let me know if I've made a mistake. Note that a is the derivative of v, which is the derivative of r, and G,M, and m are constants.

Start with $F=ma$
The gravitational force between two objects is $$F_g=-\frac{GMm}{r^2},$$ so
$$-\frac{GMm}{r^2}=ma$$ and $$-\frac{GM}{r^2}=a.$$
Now, $$a=\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr},$$ so
$$-\frac{GM}{r^2}=v\frac{dv}{dr}$$
$$-\frac{GM}{r^2}dr=vdv$$
Integrating both sides I get
$$\int_{r_o}^{r}-\frac{GM}{r^2}dr=\int_{v_o}^{v}vdv$$ where$r_o$ and $v_o$ are initial radius and velocity. This becomes$$\frac{GM}{r}-\frac{GM}{r_o}=\frac{1}{2}(v^2-v_o^2)$$
$$\pm\sqrt{\frac{2GMr_o-2GMr+v^2_or_or}{rr_o}}=\frac{dr}{dt}$$
$$\pm\int_{r_0}^{r}\sqrt{\frac{r_or}{2GMr_o-2GMr+v^2_or_or}}dr=\int_{0}^{t}dt$$
$$\pm\int_{r_0}^{r}\sqrt{\frac{r_or}{2GMr_o-2GMr+v^2_or_or}}dr=t$$
Which is obscene.

I just have no idea how to do this. I've tried rewriting it all sorts of ways with little to no luck. What can I do?
 A: $$\ \int_{r_0}^{r}\sqrt{\frac{r_0r}{2GMr_0-2GMr+v_0^2r_0r}}dr=$$
$$\ =\sqrt{\frac{r_0}{v_0^2r_0-2GM}}\int_{r_0}^{r}\sqrt{\frac{r}{\frac{2GMr_0}{v_0^2r_0-2GM}+r}}dr$$
Where I assumed that $\ v_0^2r_0>2GM$. Now set
$$\ \beta=\sqrt{\frac{r_0}{v_0^2r_0-2GM}}$$
$$\ \alpha=\frac{2GMr_0}{v_0^2r_0-2GM}$$
So you need to deal with this kind of integral:
$$\ \beta\int_{r_0}^{r}\sqrt{\frac{r}{\alpha+r}}dr$$
Now put 
$$\ \sqrt{\frac{r}{\alpha+r}}=s \implies \frac{r}{\alpha+r}=s^2$$
Then 
$$\ \frac{\alpha}{(\alpha+r)^2}dr=2sds$$
Moreover:
$$\ r=(\alpha+r)s^2\implies r=\frac{\alpha s^2}{1-s^2}\implies dr=\frac{2\alpha s}{(1-s^2)^2}$$
So the integral is:
$$\ \beta \int_{\phi(r_0)}^{\phi(r)}s\cdot \frac{2\alpha s}{(1-s^2)^2}ds=$$
$$\ 2\alpha \beta \int_{\phi(r_0)}^{\phi(r)} \frac{s^2}{(1-s^2)^2}ds$$
Now in this form it should be easier
A: From
$$2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) = \dot{r}^2-v_0^2,$$
set $r = 1/u$, (this is the most important thing to remember for problems with $1/r^2$ forces) then
$$ \dot{r} = -\frac{\dot{u}}{u^2},$$
so
$$2GM\left(u_0-u\right) = \frac{\dot{u}^2}{u^4} + v_0^2. $$
Then, rearranging,
$$ 1 = \frac{\dot{u}^2}{u^4(v_0^2+2GM(u_0-u))} $$
Now you just have to find
$$ \int \frac{dx}{x^2\sqrt{1-x}} $$
and rescale to get the right integral.
(Hint: the denominator factorises as a difference of two squares, believe it or not. Or you can just do a hyperbolic substitution.)
