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Integration by substitution

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Integrate [tex]\int[/tex][tex]\frac{dz}{1+e^z}[/tex] by substitution

    2. Relevant equations



    3. The attempt at a solution

    I chose u=(1+[tex]e^{z}[/tex]) so du/dz=[tex]e^{z}[/tex] and dz=du/[tex]e^{z}[/tex].

    Therefore, [tex]\int[/tex][tex]\frac{1}{u}[/tex] [tex]\frac{du}{e^{z}}[/tex]

    I plug z=ln(u-1) in for z, so [tex]\int[/tex][tex]\frac{1}{u}[/tex] [tex]\frac{du}{u-1}[/tex]

    From here though I don't know how to integrate. Can anyone help me with the next step?
     
  2. jcsd
  3. Nov 10, 2009 #2

    Mark44

    Staff: Mentor

    Rewrite 1/(u(u -1)) as a sum: A/u + B/(u - 1). Solve for A and B so that the two expressions are identically equal. This is called partial fractions decomposition.
     
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