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    find the integral of tan^-1 x
    using the substitution method x = tan u

    1) i have differentiated x to get dx/du = sec ^2 x
    2) I have rearranged dx = sec^2x du
    3) I have replaced the dx in the equation with sec^2x du to get tan^-1x sec^2x du
    4) i have then replaced the x with tan u to get (integral sign) tan^-1 (tan u) sec^2 (tanu) du
    5) I have then simplified this to (integral sign) u sec^2(tan u) du

    is this right so far? and if so what do i do need, i'm stuck on how to integrate sec^2 (tan u)
    i think you do it by integration by parts? but I still don't know what to do.
    Please help!
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    I am not too sure of going down this route.

    Have you tried another way at all?

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    Try:  \displaystyle \int (1) \mathrm{arctan}(x) \, dx.




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    (Original post by simon0)
    I am not too sure of going down this route.

    Have you tried another way at all?

    Spoiler:
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    Try:  \displaystyle \int (1) \mathrm{arctan}(x) \, dx.





    please could you explain where you have got that from?
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    (Original post by EmLo12)
    please could you explain where you have got that from?
    Essentially you are integrating:

    arctan(x) = 1 . arctan(x), (or 1.tan^(-1)(x)).

    Using "integration by parts" should help you.

    Can you go further here?
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    (Original post by EmLo12)
    please could you explain where you have got that from?
    Your formula book gives you the derivative of arctan(x) so the obvious thing to do here is to use parts and see where that gets you.
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    (Original post by simon0)
    Essentially you are integrating:

    arctan(x) = 1 . arctan(x), (or 1.tan^(-1)(x)).

    Using "integration by parts" should help you.

    Can you go further here?
    I've done it thanks! I was a little confused because i haven't learnt about arctan yet
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    (Original post by EmLo12)
    I've done it thanks! I was a little confused because i haven't learnt about arctan yet
    :-)

    I should have also added inverse tan as well as arctan(x) (and tan^(-1)(x)).
 
 
 
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