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Find this integration

  1. Oct 6, 2013 #1


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    1. The problem statement, all variables and given/known data
    [itex]\displaystyle \int_0^1 \dfrac{xe^{tan^{-1}x}}{\sqrt{1+x^2}} dx [/itex]

    2. Relevant equations

    3. The attempt at a solution
    Let tan^-1 (x) = t
    x = tant
    The integral then reduces to

    [itex]\displaystyle \int_0^{\pi/4} tante^tdt [/itex]

    Applying integration by parts by taking tant as 1st function

    [itex]tant e^t - \displaystyle \int sec^2te^t dt [/itex]

    This has made the problem more complicated instead of simplifying.
  2. jcsd
  3. Oct 6, 2013 #2
    $$\frac{d}{dx}\left(\arctan(x)\right) ≠ \frac{1}{\sqrt{1+x^2}}$$
    To use the substitution, multiply and divide by ##\sqrt{1+x^2}##.
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