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Math Help - what is antideriv. of arctan?

  1. #1
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    what is antideriv. of arctan?

    hey,

    I've got this problem:

    [indefinite integral](1/(t^3+t))dt

    I've used the partial fractions to split the integral into 2 separate parts, which becomes:

    ln|t| + [indefinite integral](t/(t^2 + 1))dt + C

    here I used the integration by parts for the bolded integral, which gives me:

    ln|t| + t*arctan(t) + [definite integral](arctan(t))dt + C

    and I cannot figure out what is the antiderivative of arctan(t) is?


    thanks for the tips,
    Tuugii
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  2. #2
    Senior Member polymerase's Avatar
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    \int \frac{1}{t}-\frac{t}{t^2+1}=\int \frac{1}{t}-\frac{1}{2}\int \frac{(t^2+1)'}{t^2+1}=ln|t|-\frac{1}{2}ln|t^2+1|

    no arctan involved just regualr subtitution
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  3. #3
    Junior Member Pinsky's Avatar
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    i don't know how to solve arctg, but why didn't u use a subtitution
    t^2+1=u
    That would make the integral solvable.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    As the previous posts say, dealing with arctangent is not an issue here. however, if you are still curious, we would find that integral using integration by parts, with u = \arctan x and dv = 1
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Tuugii View Post
    I've used the partial fractions to split the integral into 2 separate parts,
    you dont even need to use that for this question, check this out

    \int \frac{1}{t(t^2+1)}= \frac{1+t^2-t^2}{t(t^2+1)}= \int \frac{1+t^2}{t(t^2+1)}-\frac{t^2}{t(t^2+1)}=\int \frac{1}{t}-\frac{t}{t^2+1}
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  6. #6
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    thanks a lot for the replies.

    was really helpful!
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