# Evaluating indefinite integral - toughie!

1. Oct 27, 2005

### don_anon25

Evaluating indefinite integral -- toughie!

I have the velocity function v(x) = [(k*x^2)/(2*m)] + v0
I need to integrate this to get position as a function of time.
So v = dx/dt.
Separating variables, I get t = Integral [2m/(2mv0 + kx^2)]
Here's where I'm stuck...If i pull out the 2m, then I get something that resembles the integral which equals arc tangent.
Am I on the right track?
Can someone check my though process?

2. Oct 27, 2005

### Jameson

I assume m and k are constants. If they are, this integral seems easy.

$$x= \frac{k}{2m}\int x^2 +v_{0}dx$$

EDIT: Mistake. Sorry. Look at the below post.

Last edited: Oct 27, 2005
3. Oct 27, 2005

### HallsofIvy

Staff Emeritus
Jameson: You understand, don't you, that your response makes no sense? You have x on one side and an integral with respect to x on the other! What happened to t?

We are given
$$v(x)= \frac{dx}{dt}= \frac{kx^2}{2m}+ v_0$$
That gives, as don_anon25 says
$$\int\frac{dx}{\frac{kx^2}{2m}+ v_0}= \int dt$$
Yes, that does look like an arctangent. What's wrong with that?
If we factor out v0 on the left, we get
$$\frac{1}{v_0}\int\frac{dx}{\frac{k}{2mv_0}x^2+ 1}= t+ C$$
Now make the substitution
$$u= \sqrt{\frac{k}{2mv_0}}x$$
and the integral on the left becomes
$$\sqrt{\frac{k}{2mv_0^3}}\int{\frac{du}{u^2+ 1}$$
which is, indeed, an arctangent.

Last edited: Oct 27, 2005
4. Oct 27, 2005

### Jameson

Sorry bout that. Wasn't thinking obviously. Thanks for the correction :)