plz integrate

how do u integrate 1+u/3u+2u^2

please help me get started ...tnk u

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asked by allison
  1. I would do partial fractions.

  2. assume you mean
    (1+u) du /(3u+2u^2)
    which is
    (1+u) du/[u(3+2u)]
    which is
    1 du/[u(3+2u)] + u du/[u(3+2u)]
    now let's see if we can expand 1/[u(3+2u)]
    into
    a/u + b/(3+2u) = 1/ [u(3+2u)]
    a(3+2u) + b(u) = 1
    3 a = 1 so a = 1/3
    2a + b = 0 so b = -2/3
    so we have
    1/[u(3+2u)] = (1/3)/u - (2/3)/(3+2u)
    so the integral now is
    (1/3)du/u -(2/3)du/(3+2du) +du/(3+2u)
    or
    (1/3)du/u +(1/3) du/(3+2u)
    ok from there?

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    posted by Damon

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