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Math Help - Calc help intergration

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    Calc help intergration

    can someone tell me how the 2+8 got there and the 4 turned into a du??


     2\int 1+ \frac{4}{u^2 - 4} du= 2+8 \int \frac{du}{(u+2)(u-2)}
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    Quote Originally Posted by Legendsn3verdie View Post
    can someone tell me how the 2+8 got there and the 4 turned into a du??


    2\int 1+ \frac{4}{u^2 - 4} du= 2+8 \int \frac{du}{(u+2)(u-2)}
     <br />
2\int \left( 1+ \frac{4}{u^2 - 4} \right) du=2\int 1du+2\int \frac{4}{u^2-4}du=

    2\int du +8\int\frac{1}{u^2-4}du=2\int du +8\int\frac{du}{(u-2)(u+2)}
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    Quote Originally Posted by TheEmptySet View Post
     <br />
2\int \left( 1+ \frac{4}{u^2 - 4} \right) du=2\int 1du+2\int \frac{4}{u^2-4}du=

    2\int du +8\int\frac{1}{u^2-4}du=2\int du +8\int\frac{du}{(u-2)(u+2)}


    hm can u explain with words what is going on i m still a bit not sure what happened there.
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    Quote Originally Posted by Legendsn3verdie View Post
    can someone tell me how the 2+8 got there and the 4 turned into a du??


     2\int 1+ \frac{4}{u^2 - 4} du= 2+8 \int \frac{du}{(u+2)(u-2)}

    could it just because of this.. tell me what you guys think..

    2(1+4)\int \frac{du}{u^2-4}


    ^if this is valid.. that means i pulled a (1+4) across the intergral sign.. is that possible?
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    Quote Originally Posted by Legendsn3verdie View Post
    can someone tell me how the 2+8 got there and the 4 turned into a du??


     2\int 1+ \frac{4}{u^2 - 4} du= 2+8 \int \frac{du}{(u+2)(u-2)}
    this formula is not correct it is missing an integral sign after the 2.
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    Quote Originally Posted by TheEmptySet View Post
    this formula is not correct it is missing an integral sign after the 2.
    very strange that is wrong.. i got it directly from the solution manual



    well here is the orignal problem...

    \int \frac{x^{1/2}}{x-4}dx
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    Quote Originally Posted by Legendsn3verdie View Post
    well here is the orignal problem...

    \int \frac{x^{1/2}}{x-4}dx
    When you did your u sub u=x^{1/2}

    \int \frac{x^{1/2}}{x-4}dx=2\int \frac{u^2}{u^2-4}du

    You had this, but

    2\int \frac{u^2}{u^2-4}du=2\left( \int 1du +\int \frac{4}{u^2-4}du\right)

    Now we distribute the 2 to get

    2\int du + 2\int \frac{4}{u^2-4}du

    pulling the four out in front of the 2nd integral gives

    2\int du + (2)4\int \frac{1}{u^2-4}du

    factoring the denominator of the 2nd gives

    2 \int du + 8 \int \frac{1}{(u-2)(u+2)}du

    Note that these are the same thing

    8 \int \frac{1}{(u-2)(u+2)}du=8 \int \frac{du}{(u-2)(u+2)}

    I hope this clears it up.

    Good luck.
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  8. #8
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    Quote Originally Posted by TheEmptySet View Post
    When you did your u sub u=x^{1/2}

    \int \frac{x^{1/2}}{x-4}dx=2\int \frac{u^2}{u^2-4}du

    You had this, but

    2\int \frac{u^2}{u^2-4}du=2\left( \int 1du +\int \frac{4}{u^2-4}du\right)

    Now we distribute the 2 to get

    2\int du + 2\int \frac{4}{u^2-4}du

    pulling the four out in front of the 2nd integral gives

    2\int du + (2)4\int \frac{1}{u^2-4}du

    factoring the denominator of the 2nd gives

    2 \int du + 8 \int \frac{1}{(u-2)(u+2)}du

    Note that these are the same thing

    8 \int \frac{1}{(u-2)(u+2)}du=8 \int \frac{du}{(u-2)(u+2)}

    I hope this clears it up.

    Good luck.
    ty sir wish i had more reputation power to give you.
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