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Math Help - substitution integration

  1. #1
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    substitution integration

    hmmm...I have been studying some of the internet sites on this subject and now I think I have confused myself even more...this is an independent study class and the textbook shows very few examples...

    integral (y^2)/(y^3+5)^2 dy

    u=y^3 +5
    du = 3y^2
    dx = y^2?

    y/u integral du/3y^2 dx

    y/3y^2 ln[u] +c

    y/3y^2 ln [y^3+5] + C
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  2. #2
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    Quote Originally Posted by startingover View Post
    hmmm...I have been studying some of the internet sites on this subject and now I think I have confused myself even more...this is an independent study class and the textbook shows very few examples...

    integral (y^2)/(y^3+5)^2 dy

    u=y^3 +5
    du = 3y^2
    dx = y^2?
    \int \frac{y^2}{(y^3+5)^2} dy

    Let x=y^3+5 \Rightarrow x'=3y^2

    Thus,

    \frac{1}{3} \frac{1}{x^2} dx

    -\frac{1}{3} \cdot \frac{1}{x} +C

    -\frac{1}{3}\cdot \frac{1}{y^3+5} +C
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post
    hmmm...I have been studying some of the internet sites on this subject and now I think I have confused myself even more...this is an independent study class and the textbook shows very few examples...

    integral (y^2)/(y^3+5)^2 dy

    u=y^3 +5
    du = 3y^2
    dx = y^2?

    y/u integral du/3y^2 dx

    y/3y^2 ln[u] +c

    y/3y^2 ln [y^3+5] + C
    it seems you have confused yourself.

    \int \frac {y^2}{\left( y^3 + 5 \right)^2}dy

    Let u = y^3 + 5

    \Rightarrow du = 3y^2 dy

    \Rightarrow \frac {1}{3}du = y^2 dy

    So our integral becomes:

    \frac {1}{3} \int \frac {1}{u^2} du = \frac {1}{3} \int u^{-2}du

    = - \frac {1}{3} u^{-1} + C

    = - \frac {1}{3u} + C

    = - \frac {1}{3 \left( y^3 + 5 \right)} + C


    EDIT: You're too fast for me TPH
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  4. #4
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    how did we get -1/3?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post
    how did we get -1/3?
    By the power rule:

    \int x^{n} dx = \frac {x^{n+1}}{n + 1} + C

    So, \int u^{-2}du = \frac {u^{-2 + 1}}{-2 + 1} + C = - \frac {1}{u} + C

    we had the \frac {1}{3} already, so we made it negative because of the minus sign we get from the integral of u^{-2}
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