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Math Help - Simple Cal 2 integral

  1. #1
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    Simple Cal 2 integral

    integral of \int \,  2x/x-4<br />
    got a test tomorrow and I have a feeling this may be on the test.

    help please.
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  2. #2
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    Quote Originally Posted by p00ndawg View Post
    integral of 2x/x-4<br />
    got a test tomorrow and I have a feeling this may be on the test.

    help please.
    rewrite as follows or use long division

    2\int \frac{x}{x-4}dx=2\int \frac{(x-4)+4}{x-4}dx=

    2\int dx + 2\int \frac{4}{x-4}dx=2x+8 \ln|x-4|+c
    Last edited by TheEmptySet; March 27th 2008 at 07:23 PM. Reason: abs Thanks krizalid
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    rewrite as follows or use long division

    2\int \frac{x}{x-4}dx=2\int \frac{(x-4)+4}{x-4}dx=

    2\int dx + 2\int \frac{4}{x-4}=2x+8 \ln(x-4)+c
    Brilliant.

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  4. #4
    Junior Member teuthid's Avatar
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    This is in essence exactly the same procedure that the others have used, but with slightly different symbols (my old calc teacher used to call it "tricky" substitution)...

    choose u=x-4. This means that du=dx, but it also means that u+4=x. As a result we have the following integral:

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

    <br />
=\int \frac{2u}{u} +\frac{8}{u} du

    <br />
=2\int du + 8\int \frac{1}{u} du

    =2u+8\ln{u}+c

    =2(x-4)+8\ln{|x-4|}+c

    =2x+8\ln{|x-4|}+C
    Last edited by teuthid; March 27th 2008 at 06:44 PM. Reason: humble submission to Krizalid's superior integration skills
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  5. #5
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    Quote Originally Posted by teuthid View Post
    This is in essence exactly the same procedure that the others have used, but with slightly different symbols (my old calc teacher used to call it "tricky" substitution)...

    choose u=x-4. This means that du=dx, but it also means that u+4=x. As a result we have the following integral:

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

    <br />
=\int \frac{2u}{u} +\frac{8}{u} du

    <br />
=2\int du + 8\int \frac{1}{u} du

    =2u+8\ln{u}+c

    =2(x+4)+8\ln{(x-4)}+c

    =2x+8\ln{(x-4)}+C

    what happens to the 8 when you distribute 2 to x + 4 and why is it x + 4, shouldnt it be x - 4?
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  6. #6
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    Krizalid's Avatar
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    TheEmptySet & teuthid are missing something important here: it's actually \ln|x-4|.
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  7. #7
    Junior Member teuthid's Avatar
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    your right, it should be x-4. sorry about the typo. The 8 got dumped into the arbitrary constant that is added to the end. Any time you have an antiderivative with spare constants lying around you can do that. For example if your antiderivative was F(x)=x^2+1+\sin(x)+2+e^x+5, then the indefinite integral could be written as x^2+\sin(x)+e^x+C. You'll notice that I switched from c to C to show this "dumping".
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