# Anti-differentiation Question!

1. Jul 23, 2009

### EskShift

1. The problem statement, all variables and given/known data
If
dy / dx = 3 / (2x - 1)^3/2
and c is a real constant, then y is equal to...

2. Relevant equations
I figured since it is somewhat like the equation 1 / ax + b, then to work it out i must re-work the equation: 1/a LOGe |(ax + b)| + c

3. The attempt at a solution
i have done:
3 $$\int$$ 1 / (2x - 1)^3/2 dx
= 3/2 LOGe (2x - 1) ^3/2 + c

But not sure what the next step is? Do i get rid of the LOGe and the ^3/2?

2. Jul 23, 2009

### jgens

If I'm reading your formula correctly, I think you're anti-differentiating this function wrong. This can be handled simply with a u-substitution: Let u = 2x - 1 and go from there.

Edit: This integration is not like 1/(ax + b), it is of the form 1/(ax + b)3/2. Do you understand the difference?

3. Jul 23, 2009

### EskShift

Yeah i know what you mean, im just a little confused as to where the 3/2 goes? Its a multiple choice question and all the answers are very different to what i keep getting.

4. Jul 23, 2009

### jgens

That's because you're anti-differentiating the expression incorrectly. Once you use the u-substitution I suggested, this problem becomes a trivial use of the power rule. You should have:

dy/dx = 3/u3/2

Now, from the chain rule, we have that:

dy/du * du/dx = 3/u3/2

Can you solve the problem now?

5. Jul 23, 2009

### EskShift

I think so, i got:
-9 / (2x - 1)5/2 + c

That sound right or did i miss something?

6. Jul 23, 2009

### jgens

Well, it almost looks like you just differentiated the function again which is certainly not correct!

Hint: if y = xn then dy/dx = nxn-1 . . . if dy/dx = xn then y = xn+1/(n + 1)

7. Jul 23, 2009

### EskShift

im still not understanding this, can you please explain how to do it?

8. Jul 23, 2009

### jgens

Unfortunately, no. These forums are not here to do your homework for you, they are here to help you. So here's more help! If there's anything you're not clear on, don't hesitate to ask!

If u = 2x - 1 then du/dx = 2. Then, by the chain rule we have that 2 * dy/du = 3/u3/2. This means that 2y = 3∫u-3/2 du. Can you do the problem from here?

9. Jul 23, 2009

### EskShift

Ok so thought du/dx was 2, and so its dy/du * 2 = 3/u3/2.
That was where i got stuck :P, i thought, well then dy/du must equal 3 / u3/2 / 2
But as you just mentioned, i left out another du.
so its 2y = 3∫u3-/2 * du
therefore 2y = 3∫u3-/2 * 2 (du = 2 right?)
can just you just confirm thats correct?

10. Jul 23, 2009

### jgens

Your first thoughts were actually closer to the mark! In this case, du != 2 especially since du = 2dx. The du following the integral sign essentially just means that you are anti-differentiating the expression with respect to u rather than x or some other variable.

11. Jul 23, 2009

### EskShift

ohh i understand. ok well back to my first thoughts, in a way i think i can write it:
y = (3 ∫u ^-3/2) / 2
y = (3 ∫u ^-1/2) / 2 * -1/2
y = (3 ∫u ^-1/2) / -1
if thats correct a hint for the next step please?

12. Jul 23, 2009

### jgens

Other than sloppy notation, that looks almost right! Once you've anti-differentiated a function you no longer include the integral sign and it appears that you've forgotten the constant of integration.

13. Jul 23, 2009

### EskShift

i dont know if i have really learnt all this yet, to me its a really tough question so im assuming im not quite there yet lol.
but none-the-less, the answer i recieved on paper was:
-6 / (2x - 1)^1/2 + c
but from the way i have been going its probably wrong...

14. Jul 23, 2009

### jgens

Well, that answer look right other than the fact that it includes an extra factor of 2. How did you get the extra factor of 2?

15. Jul 23, 2009

### EskShift

ohh no! i was looking at the wrong equation. mistake on my behalf, i would say its supposed to be the same equation but with -3 at the top?
that looks better in my eyes but your opinion would be great!

16. Jul 23, 2009

### jgens

Yup, the answer I got is y = C - 3(2x - 1)-1.