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Math Help - difficult integral

  1. #1
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    difficult integral

    Hi everyone,

    Can you tell me if this is correct, please?

    Evaluate the following integral:

    x/(1+x^4 )dx

    u=x/(1+x^4)
    du=arctanx dx

    x/1+x^4=
    udu=
    (x/(1+x^4))arctanx+c

    Thank you very much
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Can you tell me if this is correct, please?

    Evaluate the following integral:

    x/(1+x^4 )dx

    u=x/(1+x^4)
    du=arctanx dx

    x/1+x^4=
    udu=
    (x/(1+x^4))arctanx+c

    Thank you very much
    i'm afraid that's incorrect. here's one way:


    \int \frac {x}{1 + x^4}~dx = \int \frac {x}{1 + \left( x^2 \right)^2}~dx

    We proceed by substitution

    Let u = x^2

    \Rightarrow du = 2x~dx

    \Rightarrow \frac {1}{2} du = x~dx

    So our integral becomes:

    \frac {1}{2} \int \frac {1}{1 + u^2}~du

    = \frac {1}{2} \arctan u + C

    = \frac {1}{2} \arctan \left(x^2 \right) + C




    Allow me to stress that \int \frac {1}{1 + x^2}~dx = \arctan x + C NOT \frac {d}{dx} \left( \frac {1}{1 + x^2} \right) = \arctan x

    and you didn't even have that, you had \frac {d}{dx} \left( \frac {x}{1 + x^4}\right) = \arctan x now that's just wrong
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    \int {\frac{x}<br />
{{1 + x^4 }}\,dx} = \frac{1}<br />
{2}\int {\frac{{\left( {x^2 } \right)'}}<br />
{{1 + \left( {x^2 } \right)^2 }}\,dx} = \frac{1}<br />
{2}\arctan x^2 + k,\,k\in\mathbb R

    Jhevon, don't be so hard, not necessary to rub her(his) in, the "not"
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Jhevon, don't be so hard, not necessary to rub her(his) in, the "not"
    it wasn't my intention to be harsh.

    but i bet he/she will never forget now

    it's just tough love
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  5. #5
    Member
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    Thank you very much

    Regards
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