# Odd integration problem

1. Oct 6, 2013

### MuIotaTau

1. The problem statement, all variables and given/known data

In the process of solving a physics problem, I've run across an integration that I'm not sure I've ever had to perform before. It's of the form $$\int \frac{1}{a - (\frac{dy}{dx})(x)} dx$$ where $a$ is a constant.

2. Relevant equations

Table integral?

3. The attempt at a solution

I honestly have no clue where to even begin with this. I do not have a functional form for $y$, it's simply a quantity that changes with respect to $x$. Any hints on where to begin?

2. Oct 6, 2013

### Ray Vickson

Your notation is confusing: is the denominator supposed to be $1 - a x y'(x)$ or $1-a y'(x)$, where $y'(x) = dy(x)/dx.$ That is, is the (x) an argument of f', or does it multuply f'(x)?

As far as I can see you can write more-or-less ANY integrand h(x) in that form: just put
$$h(x) = \frac{1}{a-y'(x)} \Longrightarrow y'(x) = \frac{h(x)-1}{a h(x)}$$ if your integrand is 1/(1-ay'), or put
$$h(x) = \frac{1}{a - x y'(x)} \Longrightarrow y'(x) = \frac{1-h(x)}{a x h(x)}$$

So, to summarize: your integrand can be anything at all, so there is no way to do the integral.

3. Oct 6, 2013

### MuIotaTau

I'm sorry, $x$ is being multiplied in the denominator. Well, um, that's certainly disappointing. I guess it's back to the drawing board. Thank you for your help!

4. Oct 6, 2013

### Dick

If it's a physics problem you probably know something about the relation between y' and x. What kind of problem is it?

5. Oct 6, 2013

### MuIotaTau

Momentum transport. In this case, $x$ is time, and $y'(x)$ is the change in mass with respect to time. So really the integral is $$\int_0^t \bigg(\frac{1}{M + m - (t')(\frac{dm}{dt'})}\bigg) dt'$$ where $t'$ is a dummy variable. $M$ and $m$ are the masses of a truck and of an original mass of sand, respectively, so they're constants, and $(t)(\frac{dm}{dt})$ is the amount of sand that falls out the truck after a time $t$.

6. Oct 6, 2013

### Dick

Since you are multiplying t' times dm/dt', I think it's pretty likely you are assuming the rate of mass change dm/dt' is constant. Are you?

7. Oct 6, 2013

### MuIotaTau

Oh no, you're right, I certainly am! So I should be able to manipulate that into something of the form $$\int \frac{1}{1 + x} dx$$ right? Which is just $ln(1 + x)$?

Last edited: Oct 6, 2013
8. Oct 6, 2013

### Dick

Sure, so the integral isn't a real problem. Now I'd worry that the integral is computing something like the average value of 1/mass over a portion of the trip, which doesn't sound very useful for momentum problems, but then I don't know exactly what you are doing.

9. Oct 6, 2013

### MuIotaTau

I'm actually calculating the impulse and dividing by the mass to get velocity, but the force in this particular case happens to be time independent, so I just pulled it out of the integral.

10. Oct 6, 2013

### Dick

Ok, carry on.