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Math Help - Integration Problem

  1. #1
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    Integration Problem

    \displaystyle \int \frac{2 x^2 + 1}{x^2}\, dx

    What technique do I use to figure this out? I know the answer, but I can't figure out how to arrive at it!

    I am trying method of substitution:
    u=2x^2+1
    du=4xdx

    I don't see any possible substitutions...
    I then tried:
    x=\sqrt{\frac{u-1}{2}}

    when I put this back into the original I have:
    \frac{2u*du}{u-1}

    now, I am not sure where to go from here.. Am I on the right track?

    Thanks!
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  2. #2
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    \displaystyle\frac{2x^2+1}{x^2}=2+\frac{1}{x^2}
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  3. #3
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    Quote Originally Posted by Vamz View Post
    \displaystyle \int \frac{2 x^2 + 1}{x^2}\, dx

    What technique do I use to figure this out? I know the answer, but I can't figure out how to arrive at it!

    I am trying method of substitution:
    u=2x^2+1
    du=4xdx

    I don't see any possible substitutions...
    I then tried:
    x=\sqrt{\frac{u-1}{2}}

    when I put this back into the original I have:
    \frac{2u*du}{u-1}

    now, I am not sure where to go from here.. Am I on the right track?

    Thanks!
    \displaystyle \frac{2x^2+1}{x^2} = 2 + x^{-2}

    now integrate ...
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  4. #4
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    Thank you very much I got it.

    I have one more question:

    \int x^5e^{x^6}

    I was able to solve this by method of staring & got \frac{e^{x^6}}{6}

    I tried to apply the method of substitution to this for kicks, and heres what happens:

    u=e^{x^6}
    du=6x^5e^{x^6}dx

    As a general rule for the future, what exactly does it mean when my du contains my u? What should I do in these scenarios?

    <br />
\int \frac{du}{6} = \frac{1}{6}\int du<br />

    Would this be the proper way to proceed?
    When all I have is my du in my expression, does that always mean to stop "integrating" and to just write it as-is?

    Thanks!
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  5. #5
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    Quote Originally Posted by Vamz View Post
    Thank you very much I got it.

    I have one more question:

    \int x^5e^{x^6}

    I was able to solve this by method of staring & got \frac{e^{x^6}}{6}

    I tried to apply the method of substitution to this for kicks, and heres what happens:

    u=e^{x^6}
    du=6x^5e^{x^6}dx

    As a general rule for the future, what exactly does it mean when my du contains my u? What should I do in these scenarios?

    <br />
\int \frac{du}{6} = \frac{1}{6}\int du<br />

    Would this be the proper way to proceed?
    When all I have is my du in my expression, does that always mean to stop "integrating" and to just write it as-is?

    Thanks!
    u= x^6 not e^{x^6}
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