Integrate (X/X+d)dX: Solve w/Substitution

[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

 

[gnome]

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

 

[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.

 

[Hurkyl]

but the num and den degrees are the same

What does that have to do with anything?

 

[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??

What does that have to do with anything?

You're very helpful

 

[Hurkyl]

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)

 

[TheDestroyer]

Very Simple Way, Here It,

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

then the answer will be:

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

hehe, isn't that easy?

 

[mewmew]

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

 

[TheDestroyer]

HEHE, right, i fixed it :P

 

[funkwort]

Very Simple Way, Here It,

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

then the answer will be:

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

hehe, isn't that easy?

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).

 

[Hurkyl]

I don't understand why the below procedure doesn't work when it's just another simple substitution??

Because you forgot the constant of integration.

 

[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?

 

[Hurkyl]

Shouldn't they produce the same value considering they are merely two different substitutions?

The two answers are the same.

 

[funkwort]

no ... x does not equal x+d

you're not helping me here

 

[cepheid]

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?

 

[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

 

[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))<br /> = L - 0 + d - d - d * \ln(L+d) + d * \ln(0+d) = L - d(\ln(L+d) - \ln(d))<br /> = 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.

 

[funkwort]

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