# How to integrate [ x / (1 + x) ] dx

by Color_of_Cyan
Tags: integrate
 P: 301 1. The problem statement, all variables and given/known data Evalute the integral ∫ [x / 1 + x] dx 2. Relevant equations ∫ [x / 1 + x] dx 3. The attempt at a solution I forgot how to do solve this type of integral, or never had enough practice. And this problem is actually for a physics problem :-) And my algebra is very rusty as well. ∫ [x / 1 + x] dx Can't do U-sub, don't think you can do by parts, can't do trig substitution, not sure about partial fractions Divide everything by 'x'. ∫ [x / 1 + x] dx = ∫ [x (1 / x)/ 1(1 / x) + x(1 / x)] dx ∫ [1 / (1/x) + (1)] dx Not sure where to go from here.
 HW Helper Thanks PF Gold P: 7,632 Try writing $$\frac x {1+x}=1 -\frac 1 {1+x}$$
 P: 301 Yes, I see that will work, but I just can not see a straightforward way to get there :-(
 Emeritus Sci Advisor HW Helper PF Gold P: 7,808 How to integrate [ x / (1 + x) ] dx ... to get there from where? Can you find $\displaystyle \int\,\left(1 -\frac 1 {1+x}\right)\,dx\ ?$
HW Helper
Thanks
PF Gold
P: 7,632
 Quote by Color_of_Cyan Yes, I see that will work, but I just can not see a straightforward way to get there :-(
It is a standard procedure when you have a polynomial of equal or higher degree in the numerator to reduce it by long division until the numerator is less degree than the denominator. Just divide x+1 into x long division.
P: 867
 Quote by Color_of_Cyan Yes, I see that will work, but I just can not see a straightforward way to get there :-(
There's a trick that's often used for problems like this:

$$\frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = \frac{x + 1}{x + 1} - \frac{1}{x + 1} = 1 - \frac{1}{x + 1}$$

You could also do this easily with the substitution u = x + 1 → u - 1 = x and du = dx. After the substitutions, then you can split up the integrand into two fractions, which could be a little easier than the above method.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,808 Let's wait to hear from Cyan before we do any more. We're not sure what his/her hang up is.
P: 72
My main point:
The word easy should not be in a tutor's vocabulary. Although, for those who have spent hours (sometimes hundreds) mastering various Calculus material, few if any of us thought it was easy to learn.

 Quote by Bohrok You could also do this easily with the substitution u = x + 1 → u - 1 = x and du = dx. After the substitutions, then you can split up the integrand into two fractions, which could be a little easier than the above method.
The most effective method, for most students is to realize, not only u=x+1 can be substituted in, but we may replace other x's using x=u-1. I hope this didn't get lost in the mix.

I hope SammyS is Ok, with this interjection.
P: 301
 Quote by SammyS Let's wait to hear from Cyan before we do any more. We're not sure what his/her hang up is.

 Quote by Bohrok There's a trick that's often used for problems like this: $$\frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = \frac{x + 1}{x + 1} - \frac{1}{x + 1} = 1 - \frac{1}{x + 1}$$ You could also do this easily with the substitution u = x + 1 → u - 1 = x and du = dx. After the substitutions, then you can split up the integrand into two fractions, which could be a little easier than the above method.
Thanks, this really helps a lot!

The only problem is that I get a slightly different answer for the U-Substitution method you provided, probably because I'm being dumb making mistakes here and there:

∫ [ x / (x + 1) ] dx

u = x + 1 ; du = dx

x = u - 1

∫ [ (u - 1) / u ] du

= ∫ [ u / u] du - ∫ [1 / u] du

= ∫ 1 du - ∫ [1 / u] du

= (u) - ln(u) ; recall u = x + 1; = x+1 - ln (x+1)

The first method you said (and I'll copy it over):

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

= [ (x + 1) / (x + 1) ] - [ 1 / (x + 1) ]

= 1 - [1 / (x + 1)]

Integrating this now:

∫ 1 - [1 / (x + 1)] dx

= x - ln(x + 1)
 P: 72 Two things for *indefinite integrals*, always remember the constant of integration. Get all the easy points on quizzes & tests. Second, they only differ by a constant, right? So pick different constants of integration to compensate. +4 for one & +5 for the other. Very commonly when doing an integral by two different methods, the answer will come out the same, but differ by an addition constant. Thank you very much for coming back and finishing the question. Next step for you: develop more self sufficiency on questions like this.
P: 301
 Quote by nickalh Two things for *indefinite integrals*, always remember the constant of integration. Get all the easy points on quizzes & tests. Second, they only differ by a constant, right? So pick different constants of integration to compensate. +4 for one & +5 for the other. Very commonly when doing an integral by two different methods, the answer will come out the same, but differ by an addition constant. Thank you very much for coming back and finishing the question. Next step for you: develop more self sufficiency on questions like this.
Yeah I should have put the + C also, but this was for a physics problem and this integral was the last step from me finishing it, and it actually was a definite integral where I integrated over a length.
 P: 2 I realize that this is a ridiculous question, but what algebraic technique are you using? Are you completing the square?
P: 2
 Quote by John_Rambo I realize that this is a ridiculous question, but what algebraic technique are you using? Are you completing the square?
 Quote by Bohrok There's a trick that's often used for problems like this: $$\frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = \frac{x + 1}{x + 1} - \frac{1}{x + 1} = 1 - \frac{1}{x + 1}$$ You could also do this easily with the substitution u = x + 1 → u - 1 = x and du = dx. After the substitutions, then you can split up the integrand into two fractions, which could be a little easier than the above method.
I realize that this is a ridiculous question, but what algebraic technique are you using? Are you completing the square?
Emeritus