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How to integrate [ x / (1 + x) ] dxby Color_of_Cyan
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#1
Jul2311, 04:18 PM

P: 307

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 Usub, 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. 


#2
Jul2311, 04:22 PM

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Thanks
PF Gold
P: 7,719

Try writing
[tex]\frac x {1+x}=1 \frac 1 {1+x}[/tex] 


#3
Jul2311, 06:58 PM

P: 307

Yes, I see that will work, but I just can not see a straightforward way to get there :(



#4
Jul2311, 07:04 PM

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How to integrate [ x / (1 + x) ] dx
... to get there from where?
Can you find [itex]\displaystyle \int\,\left(1 \frac 1 {1+x}\right)\,dx\ ?[/itex] 


#5
Jul2311, 07:08 PM

HW Helper
Thanks
PF Gold
P: 7,719




#6
Jul2311, 08:43 PM

P: 867

[tex]\frac{x}{x + 1} = \frac{x + 1  1}{x + 1} = \frac{x + 1}{x + 1}  \frac{1}{x + 1} = 1  \frac{1}{x + 1}[/tex] 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. 


#7
Jul2311, 09:12 PM

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Let's wait to hear from Cyan before we do any more. We're not sure what his/her hang up is.



#8
Jul2411, 06:57 AM

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. I hope SammyS is Ok, with this interjection. 


#9
Jul2611, 02:13 AM

P: 307

The only problem is that I get a slightly different answer for the USubstitution 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) 


#10
Jul2611, 04:19 AM

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. 


#11
Jul2611, 11:09 AM

P: 307




#12
Feb1913, 09:20 PM

P: 2

I realize that this is a ridiculous question, but what algebraic technique are you using? Are you completing the square?



#13
Feb1913, 09:21 PM

P: 2




#14
Feb1913, 10:44 PM

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You can get same result by using long division. 


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