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Math Help - Solve inverse tan integral

  1. #1
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    Solve inverse tan integral

    Hey, I cannot figure out how to solve this integral. My teacher says we can't use the table of integrals in the back of the book, and have to solve it ourselves. Here it is:

    \int{\arctan{x}}dx

    And the answer is supposed to be:

    \int{\arctan{x}}dx=x*\arctan{x}-\ln{\sqrt{1+x^2}}+C

    Please help!!!

    Thanks
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  2. #2
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    Quote Originally Posted by DHS1 View Post
    Hey, I cannot figure out how to solve this integral. My teacher says we can't use the table of integrals in the back of the book, and have to solve it ourselves. Here it is:

    \int{\arctan{x}}dx

    And the answer is supposed to be:

    \int{\arctan{x}}dx=x*\arctan{x}-\ln{\sqrt{1+x^2}}+C

    Please help!!!

    Thanks
    integration by parts ...

    u = \arctan{x}<br />

    dv = dx

    du = \frac{1}{1+x^2} \, dx

    v = x

    \int u \, dv = u \cdot v - \int v \, du

    take it from here?
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  3. #3
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    Thank you skeeter that looks like it would work fine. However there is one problem, he did not teach us this method yet. It is only the 2nd week of this class, and all he has taught us was U-Substitution. Do you know of a way to solve it using this? Or is Integration by parts the only way?
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  4. #4
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    Quote Originally Posted by DHS1 View Post
    Thank you skeeter that looks like it would work fine. However there is one problem, he did not teach us this method yet. It is only the 2nd week of this class, and all he has taught us was U-Substitution. Do you know of a way to solve it using this? Or is Integration by parts the only way?
    Sorry, I do not know another method.
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