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Math Help - Calculus Integration

  1. #1
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    Calculus Integration

    integration (0-3) 4x/((x-5)^2)
    integration (0-inf) x^2/(4+x^6)
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  2. #2
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    Quote Originally Posted by Sally_Math View Post
    integration (0-3) 4x/((x-5)^2)
    integration (0-inf) x^2/(4+x^6)
    For the first, use u=x-5.

    For the second, do you think you can make soemthing if you know that x^6=(x^3)^2 ?
    Do not forget that the second one is an improper integral.
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  3. #3
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    i tried the u substituition but it didn't work
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  4. #4
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    Quote Originally Posted by Sally_Math View Post
    i tried the u substituition but it didn't work
    No. It works.
    Lets see your work.
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  5. #5
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    Quote Originally Posted by Sally_Math View Post
    i tried the u substituition but it didn't work
    It's not as "obvious":

    \int_0^3 \frac{4x}{(x-5)^2}dx

    u=x-5 \rightarrow u+5=x

    du=dx

    \int_0^3 \frac{4(u+5)}{u^2}du

    Can probably take it from there. . .

    For the second one, gosh - that one COULD look like the derivative of a trig function if we could only do some hand-waving (re: u-sub) to make it so. . .

    \int_0^\infty \frac{x^2}{4+x^6}dx

    Let me take u to be x^3:

    u = x^3

    du = 3x^2dx \Rightarrow \frac{du}{3} = x^2 dx

    And well hot damn, I have an "x-squared dx". Now replace all that and see what juicy derivative you get. From here it should be "elementary. . ." - hopefully. Assuming you know your trig derivatives.
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  6. #6
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    Quote Originally Posted by ANDS! View Post
    It's not as "obvious":

    \int_0^3 \frac{4x}{(x-5)^2}dx

    u=x-5 \rightarrow u+5=x

    du=dx

    \int_0^3 \frac{4(u+5)}{u^2}du

    Can probably take it from there. . .
    You should change the upper/lower limits.
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  7. #7
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    Well look at that I should. . .I'm actually going to say I left them there as a learning tool for the OP to see if he can catch mistakes - yup. That's what I did.
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