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Math Help - Substitution Integral

  1. #1
    Member Em Yeu Anh's Avatar
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    Red face Substitution Integral

    \int\frac{dx}{(1+x^{\frac{1}{4}})x^{\frac{1}{3}}}

    I am stumped on what to substitute here.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Em Yeu Anh View Post
    \int\frac{dx}{(1+x^{\frac{1}{4}})x^{\frac{1}{3}}}

    I am stumped on what to substitute here.
    EDIT: uh sorry.
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  3. #3
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    Quote Originally Posted by Em Yeu Anh View Post
    \int\frac{dx}{(1+x^{\frac{1}{4}})x^{\frac{1}{3}}}

    I am stumped on what to substitute here.
    Try x=u^{12}.
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  4. #4
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    Try to sub.  x = u^{-12}

     dx = -12u^{-13} du

    the integral goes to

     \int \frac{-12u^{-13} du}{ ( 1+ u^{-12/4}) u^{-12/3} }

     = \int \frac{-12u^{-13} du}{ ( 1+ u^{-3}) u^{-4} }

     = -12 \int \frac{du}{u^6(1 + u^3)}

     = - 12 \int \left[ \frac{1}{1+u^3} - \frac{u^3-  1}{u^6} \right] ~du

     = -12 \int  \left[  \frac{1}{1 + u^3 }  - \frac{1}{u^3} + \frac{1}{u^6}  \right] ~du

    and so on
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  5. #5
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by simplependulum View Post
    Try to sub.  x = u^{-12}

     dx = -12u^{-13} du

    the integral goes to

     \int \frac{-12u^{-13} du}{ ( 1+ u^{-12/4}) u^{-12/3} }

     = \int \frac{-12u^{-13} du}{ ( 1+ u^{-3}) u^{-4} }

     = -12 \int \frac{du}{u^6(1 + u^3)}

     = - 12 \int \left[ \frac{1}{1+u^3} - \frac{u^3- 1}{u^6} \right] ~du

     = -12 \int \left[ \frac{1}{1 + u^3 } - \frac{1}{u^3} + \frac{1}{u^6} \right] ~du

    and so on
    Thanks! that's an interesting way to solve it! would never have thought of that myself.

    Only one question,  = -12 \int \frac{du}{u^6(1 + u^3)} should that instead be  = -12 \int \frac{du}{u^9(1 + u^3)} or am I wrong?
    Last edited by Em Yeu Anh; January 25th 2010 at 10:12 PM.
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  6. #6
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    Quote Originally Posted by Em Yeu Anh View Post
    Thanks! that's an interesting way to solve it! would never have thought of that myself.

    Only one question,  = -12 \int \frac{du}{u^6(1 + u^3)} should that instead be  = -12 \int \frac{du}{u^9(1 + u^3)} or am I wrong?

    Be careful of the denominator which is  u^9(1 + u^{-3})
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  7. #7
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by simplependulum View Post
    Be careful of the denominator which is  u^9(1 + u^{-3})
    Is that not the same as what I just put?
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  8. #8
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    Quote Originally Posted by Em Yeu Anh View Post
    Is that not the same as what I just put?
    No. As an counter-example, put u=2 in both of them, and you will see that they are not equal to each other.
    In general: u^a \neq u^{-a}.
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  9. #9
    Member Em Yeu Anh's Avatar
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    My eyes completely missed that negative sign, sorry.
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