# Integrating v/(1+v²)

1. Apr 26, 2007

### Pinali

Hi. I've been struggling with this for hours and not getting anywhere even after a hint from a teacher. I have to integrate (indefinite integral) v / (1+cv²) where c is a constant.

The hint the teacher gave was that the integral of 1 / (1+v²) is tan^-1(v), but I can't see how to use this unless I use integration by parts - which yields another term which involves integrating tan^-1, which we haven't been taught so I assume I'm not expected to do and thus I'm doing something wrong.

I also attempted to do it by the "if the derivative of the bottom is the top then the solution is ln |bottom|, giving the solution 1/c ln |1+v²|.
The problem with this is that I then have to input v=dy/dx , rearrange and solve again to get the form I need (it's part of a question on solving a nonlinear autonomous ordinary differential equation) which, when I tried, was insanely complicated - well beyond the level I'm supposed to be working at.

I'm completely stumped so can anyone be of assistance here? I can describe the full problem if there's not enough information here on the nature of the problem.

2. Apr 26, 2007

### nrqed

That's correct. It's just a change of variable to $w \equiv 1+v^2$ and $v dv = dw/2$ which leads to a log, as you said.
well, what do you have to do next, more precisely? The result is a log and there is not much to do about it!

3. Apr 26, 2007

### rmira97

Solution

Hi Pinali,
Just take v^2=t 2vdv=dt.
Hence substitute and find the answer.

4. Apr 26, 2007

### Kruz87

Yeah, I don't know where arctan would come into play. The most obvious method would seem to be substitution as mentioned above. Set u= v^2 + 1, du=2v dv, du/2=v dv.

5. Apr 26, 2007

### Pinali

In order to explain what I have to do here, I'm trying to solve the autonomous nonlinear ODE y"=b+cy' where b and c are constants.
My lectures showed how to do this using the substitution v=y' and so y"=v*dv/dy. I then separated variables to get the above integral. Maybe it's just the fact that the teacher gave this hint that's throwing me off and the actual use for the hint doesn't come until later on?
I'll give it another go, as it seems like I was right to think it should be a log. Thanks!

6. Apr 26, 2007

### nrqed

if y'' = v dv/dy then the equation is

v dv/dy = b + c v so dv/dy = (b+c v)/v

How did you get to a form v/(1+cv^2)??

7. Apr 26, 2007

### D H

Staff Emeritus
A nonlinear ODE would be something like yy''=b+cy' or y''=b+cyy'. This is a good old second order linear ODE. There are lots of techniques to solve these.

The substitution v=y' is a good one, but rather than y''=v*dv/dy, why not use y''=v' instead? Then the original equation becomes v'=b+cv, which you should be able to solve easily. Now integrate v to yield y.

8. Apr 26, 2007

### Pinali

Ah sorry typing error. It meant to be y"=-b - c(y')² the original equation, not just y' by itself. And the signs wrong too, argh, I am silly.
v dv/dy = - b - c v²
dv/dy = - (b + c v²)/v
So then - § v / (b + c v²) dv = § 1 dy
- 1/b § v / (1 + d v²) dv = y + p
d is another constant that I choose to represent c/b (I used different notation at first so had the original as d and then chose c to be d/b or so), and p is constant of integration. And then it's the § v / (1 + d v²) dv part that stumped me.

9. Apr 27, 2007

### HallsofIvy

Staff Emeritus
Yes, for v= y', your differential equation becomes v'= -b-cv2. That can be written as dv/(b+cv^2)= -dx. You can simplify a bit by multiplying both sides by b to get dv/(1+ (c/b)v^2)= -b dx. That is NOT the same as the equation you first gave, vdv/(1+cv^2) since you do NOT have that "v" in the denominator. Now your teacher's suggestion that you use the fact that the anti-derivative of 1/(1+u^2) is arctan(u) is exactly what you need. Let $u=\sqrt{c/b}v$ to convert to a constant times 1/(1+u^2).

10. Apr 27, 2007

### Pinali

Ah yeah, I just managed it. I did indeed get a log which was apparently correct (it yielded the correct answer anyway when I applied the initial conditions), so thank you for the help. The arctan thing it seems was for use later on, which I also managed okay when I realised it. Thanks again!

11. Apr 27, 2007

### HallsofIvy

Staff Emeritus
I am still confused as to what you want to integrate. Originally you said vdv/(c+ v2) which will give a logarithm, later, you gave a problem which led to dv/(c+ v2) which, integrated, gives arctangent.