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Integration by Parts substitution

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    \int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt

    I'm stuck at this point. I think I need to make a substitution for the denominator, but I'm not sure how to go about doing so.
  2. jcsd
  3. Jan 30, 2010 #2


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    [tex]\int \frac{t}{1+16t^2}dt[/tex]

    f we let u=1+16t2 what is du equal to?
  4. Jan 30, 2010 #3
    It's good so far.

    You can substitute for [tex] V = {1+16t^2} [/tex], so [tex] dV = 32tdt[/tex].

    Then, you'll have [tex] \frac{1}{32} \int \frac{1}{V}dV[/tex].

    That equates to [tex] \frac {ln(|V|)}{32}[/tex].

    Multiply by -4, to obtain [tex] \frac {-ln(|V|)}{8}[/tex].

    Replace V and you're done.

    Good luck,

  5. Jan 30, 2010 #4
    Got it. Thank you both! Very helpful.
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