# Integration problem

1. Feb 24, 2005

### funkwort

I cannot figure out how to integrate (X/X+d)dX. I know that I should probably use the method of partial fractions but the num and den degrees are the same and the den cannot be broken down. I guess I could use some sort of substitution but which? please help

2. Feb 24, 2005

### gnome

try using long division to divide x by (x+d) & see what you get.

3. Feb 24, 2005

### Jameson

I'd do integration by parts.

You can write this as $$x * \frac{1}{x+d} dx$$

The second part should look familiar. (arctan)

The $$\int{u*dv} = uv - \int{v*du}$$

Let u = x and dv = $$\frac{1}{x+d}$$

Edit: I made a mistake. Sorry. For the arctan rule to apply there needs to be an x^2 in the bottom. Maybe you can still do it that way. I don't know.

Last edited: Feb 24, 2005
4. Feb 24, 2005

### Hurkyl

Staff Emeritus
What does that have to do with anything?

5. Feb 24, 2005

### funkwort

wait... if I let u = x+d then x = u-d and int (x/x+d)dx becomes
int(1-d/u)du = u - dlnu = x+d - dln(x+d). Is this correct??

6. Feb 25, 2005

### Hurkyl

Staff Emeritus
Thanks, I try.

I was trying to hint that partial fractions still applies when the numerator and denominator have the same degree. (At least, the first step of long division does)

7. Feb 25, 2005

### TheDestroyer

Very Simple Way, Here It,

(x)/(x+d) = (x+d-d)/(x+d) = (1) - (d/(x+d))

x - d*ln(x+d) + C

hehe, isn't that easy?????

Last edited: Feb 25, 2005
8. Feb 25, 2005

### mewmew

The above is correct, except it is "-" instead of "+".

9. Feb 25, 2005

### TheDestroyer

HEHE, right, i fixed it :P

10. Feb 25, 2005

### funkwort

Ok, I see, thanks. However, I don't understand why the below procedure doesn't work when it's just another simple substitution??

if I let u = x+d then x = u-d and int (x/x+d)dx becomes int(1-d/u)du = u - dlnu = x+d - dln(x+d).

11. Feb 25, 2005

### Hurkyl

Staff Emeritus
Because you forgot the constant of integration.

12. Feb 25, 2005

### funkwort

I understand that I forgot the constant of integration, but I am referring to the fact that Destroyer's procedure produces: x - dln(x+d) + C while the other procedure produces (x+d) - dln(x+d) + C. Shouldn't they produce the same value considering they are merely two different substitutions?

13. Feb 25, 2005

### Hurkyl

Staff Emeritus
The two answers are the same.

14. Feb 25, 2005

### funkwort

no ... x does not equal x+d

you're not helping me here

15. Feb 26, 2005

### cepheid

Staff Emeritus
Yes he is...he told you the answers were the same. You should have believed him. x is not x + d. But what is d? d is a constant. So did you really get a different answer from TheDestroyer's (lol at the name)? No, you did not. Both answers are valid *antiderivatives*, because when you differentiate them with respect to x, all constant terms drop out...and you end up back with the integrand. So absorb d into C...they are both constant terms. Your new C = C+d. Happy?

16. Feb 26, 2005

### funkwort

Ok wait, I just realized I made a mistake at the beginning of the thread. I should have clarified that this isn't an indefinite integral. The limits are from (0 to L) so:

1) let u = x+d then x = u-d
(x/x+d)dx = (1-d/u)du = u - dlnu = x+d - dln(x+d)

from (0 to L) = (L+d) - dln(1+L/d)

2) (x)/(x+d)dx = (x+d-d)/(x+d)dx = (1) - (d/(x+d))dx = x - dln(x+d)

from (0 to L) = L - dln(1+L/d)

am I still missing it, or are they still the same?

PS. forgive for going on and on

Last edited: Feb 26, 2005
17. Feb 26, 2005

### Moo Of Doom

Yes, they are still one and the same. When taking the definite integral, you will subtract one from the other, so:

$$g(L) - g(0) = L + d - d * \ln(L+d) - (0+d - d * \ln(0+d)) = L - 0 + d - d - d * \ln(L+d) + d * \ln(0+d) = L - d(\ln(L+d) - \ln(d)) = L - d * \ln(L/d + 1)$$

Notice how the d cancels out the -d (just like the arbitrary constant of integration, C, cancels out) when you take the definite integral.

Last edited: Feb 26, 2005
18. Feb 26, 2005

### funkwort

holy crap ... lol. I can't believe I didn't see that :surprised
thank you