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Not seeing how to integrate y/(y+1)

  1. Jan 27, 2006 #1
    Hello,

    I cannot figure out how to evaluate the following integral,

    [tex]\int {\frac{y}{{y + 1}}dy}[/tex]

    If it was [tex]y^2+1[/tex] then I see how a u-substitution of [tex]u=y^2+1[/tex] would work.

    Thanks
     
  2. jcsd
  3. Jan 27, 2006 #2

    StatusX

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    Homework Helper

    Usually with fractions like this, you'll find you can express what you have as the sum of two or more fractions with the same denominator which are easier to integrate. In this case, try the numerators y+1 and -1.
     
  4. Jan 27, 2006 #3
    As in, write y = y + 1 - 1.
     
  5. Jan 27, 2006 #4
    Ok, that is what I was having trouble seeing. Thanks

    [tex]
    \frac{y}{{y + 1}} = \frac{{y + 1}}{{y + 1}} - \frac{1}{{y + 1}} = 1 - \frac{1}{{y + 1}}
    [/tex]

    [tex]
    \int {1{\rm }dy - } \int {\frac{1}{{y + 1}}{\rm }} dy = y - \ln \left| {y + 1} \right| + C
    [/tex]
     
  6. Jan 27, 2006 #5

    0rthodontist

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    Science Advisor

    You could also substitute u = y + 1, du = dy, and get the integral of (u - 1) / u with respect to u.
     
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