U = (1 + x) and integrate

  • Thread starter confused88
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  • #1
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Main Question or Discussion Point

Hi!! I'm having trouble understanding my textbook, so can someone please explain to me how they got from


[tex]\int1/(1+x)[/tex]2 dx, with the range of the integral from 0 to x/n

to

1 - n/(x+n)


THank you So MuCh
 

Answers and Replies

  • #2
Pengwuino
Gold Member
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15


Are you having trouble with the integration or the final answer?

u substitution works fine here, let u = (1 + x) and integrate. The final answer is a simple algebraic trick, nothing spectacular.
 
  • #3
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I'm just having trouble with the integration ><. Oh wells
 
  • #4
Pengwuino
Gold Member
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You're looking for the integral of [tex]\mathop \smallint \nolimits_0^{x/n} \frac{1}{{(1 + x')^2 }}dx'[/tex] (technically you're not allowed to have your variable of integration in your bounds)

Make a substitution so that your integral now becomes [tex]\mathop \smallint \nolimits_0^{x/n} \frac{1}{{(u)^2 }}du\frac{{dx}}{{du}}[/tex] and remember to compute dx/du.

What, when you take its derivative becomes [tex]\frac{1}{{u^2 }}[/tex]? Find what that is, substitute back in for what you had set u to and you can plug in your integration limits and wala!
 

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