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Math Help - Integration Problem

  1. #1
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    Integration Problem

    I'm having a hard time doing this problem.

    (Tan^-1(x)) / 1 + x^2

    I'm know i'm gonna have to use the substitution rule

    I designated

    u = arctan

    du = 1 / (1 + x^2) dx

    From here i'm fairly stuck. Any help is appreciated
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    I'm having a hard time doing this problem.

    (Tan^-1(x)) / 1 + x^2

    I'm know i'm gonna have to use the substitution rule

    I designated

    u = arctan

    du = 1 / (1 + x^2) dx

    From here i'm fairly stuck. Any help is appreciated
    You have the derivative of the quantity

    therefore \int\frac{arctan(x)}{1+x^2}=\frac{arctan^2(x)}{2}+  C
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    I'm having a hard time doing this problem.

    (Tan^-1(x)) / 1 + x^2

    I'm know i'm gonna have to use the substitution rule

    I designated

    u = arctan

    du = 1 / (1 + x^2) dx

    From here i'm fairly stuck. Any help is appreciated
    O and if you do it your way...so if you set it up \int{u}du=\frac{u^2}{2}+C

    and remember u=arctan(x)
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  4. #4
    Moo
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    Hello,

    You're right for the beginning !

    u=\arctan(x)

    du=\frac{1}{1+x^2} dx \rightarrow dx=(1+x^2)du

    Then \int \frac{\arctan(x)}{1+x^2} dx=\int \frac{u}{1+x^2} dx=\int \frac{u}{\color{red}1+x^2} ({\color{red}1+x^2}) du =\int udu
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  5. #5
    o_O
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    ??? You're stuck? You just substitute it in ...

    {\color{red} u = \arctan x}
    {\color{blue} du = \frac{1}{1 +x^{2}} dx}

    Looking at your integral:
    \int \frac{{\color{red}\arctan x}}{{\color{blue}1 + x^{2}}} {\color{blue} dx} = \int \left({\color{red}\arctan x} \cdot {\color{blue}\frac{1}{1+x^{2}}dx}\right) = \int {\color{red} u} {\color{blue}du}
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    You're right for the beginning !

    u=\Arctan(x)

    du=\frac{1}{1+x^2} dx \rightarrow dx=(1+x^2)du

    Then \int \frac{\Arctan(x)}{1+x^2} dx=\int \frac{u}{1+x^2} dx=\int \frac{u}{\color{red}1+x^2} ({\color{red}1+x^2}) du =\int udu
    Quote Originally Posted by o_O View Post
    ??? You're stuck? You just substitute it in ...

    {\color{red} u = \arctan x}
    {\color{blue} du = \frac{1}{1 +x^{2}} dx}

    Looking at your integral:
    \int \frac{{\color{red}\arctan x}}{{\color{blue}1 + x^{2}}} {\color{blue} dx} = \int \left({\color{red}\arctan x} \cdot {\color{blue}\frac{1}{1+x^{2}}dx}\right) = \int {\color{red} u} {\color{blue}du}
    Haha...too quick for you all ....just kidding
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  7. #7
    Moo
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    At least we explained why
    We already discussed about that, it's no competition
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  8. #8
    o_O
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    Next time stud ...

    Edit: omg moo beat me again.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    At least we explained why
    We already discussed about that, it's no competition
    I explained ...I said he had \int{f(g(x))\cdot{g'(x)}}dx

    'Nuff said
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  10. #10
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    Thanks guys!
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