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Math Help - Integration by Substitution

  1. #1
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    Integration by Substitution

    just two problems that are giving me grief

    \int(x^2)+\frac{1}{(3x)^2} dx

    \int\frac{x^2+3x+7}{\sqrt{x}} dx

    thanks
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  2. #2
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    For the first one:
     \int x^2 + \frac{1}{(3x)^2}dx

    I would just split them up to make it easier like this and also just fix up that 3x^2 to make it look a bit nicer:

     \int x^2dx + \int \frac{1}{9x^2}dx

    The first part is easy...

     \int x^2dx = \frac {x^3}{3}

    The second part is a little bit harder...

     \int \frac{1}{9x^2} dx = \frac{1}{9} \int x^-2dx

    Then for the  \int x^-2dx all it is... is  \frac {x^-1}{-1}

    So for a final answer you have...
     \frac {x^3}{3} - \frac{1}{9x}
    Last edited by larson; February 27th 2008 at 05:14 PM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by larson View Post
    For the first one:
     \int x^2 + \frac{1}{(3x)^2}

    I would just split them up to make it easier like this and also just fix up that 3x^2 to make it look a bit nicer:

     \int x^2 + \int \frac{1}{9x^2}

    The first part is easy...

     \int x^2 = \frac {x^3}{3}

    The second part is a little bit harder...

     \int \frac{1}{9x^2} = \frac{1}{9} \int x^-2

    Then for the  \int x^-2 all it is... is  \frac {x^-1}{-1}

    So for a final answer you have...
     \frac {x^3}{3} - \frac{1}{9x}
    don't forget you arbitrary constant, nor the dx at the end of the integrals. otherwise, good job
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by doctorgk View Post
    \int\frac{x^2+3x+7}{\sqrt{x}} dx
    for this one, note that \sqrt{x} = x^{1/2}. divide each term by it and then you just use the power rule
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    don't forget you arbitrary constant, nor the dx at the end of the integrals. otherwise, good job
    haha, whoops... sometimes i do forget those :\, but thanks, I added them on.
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