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Math Help - Indefinite Integral

  1. #1
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    Indefinite Integral

    This is what I got
    \int(1+\frac{1}{x})^{-2}(\frac{1}{x^2})\,dx=-(x+1)^{-1}+C
    The book says the answer is (1+\frac{1}{x})^{-1}+C
    Why is my answer wrong?
    Thanks
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  2. #2
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    The actual problem would help immensely.
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  3. #3
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    The actual problem was
    \int(1+\frac{1}{x})^{-2}(\frac{1}{x^2})\,dx
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  4. #4
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    They made a different substitution than you:

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

    u = 1+\frac{1}{x}
    du = \frac{-1}{x^{2}}dx
    -\int (u)^{-2}du

    etc.
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  5. #5
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    Quote Originally Posted by ANDS! View Post
    They made a different substitution than you:

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

    u = 1+\frac{1}{x}
    du = \frac{-1}{x^{2}}dx
    -\int (u)^{-2}du

    etc.
    Thanks. I see how to get the answer in the book now, but I can't figure out what I did wrong to get a different answer.
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  6. #6
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    Quote Originally Posted by ione View Post
    Thanks. I see how to get the answer in the book now, but I can't figure out what I did wrong to get a different answer.
    If you post your working someone may be able to help you
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  7. #7
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    Quote Originally Posted by ione View Post
    Thanks. I see how to get the answer in the book now, but I can't figure out what I did wrong to get a different answer.
    Actually both solutions are the same.

    \left(1+\frac{1}{x}\right)^{-1}+C_1

    =~ \frac{1}{1+\frac{1}{x}}+C_1

    =~ \frac{x}{x+1}+C_1

    =~ \frac{x+1-1}{x+1}+C_1

    =~ 1+\frac{-1}{x+1}+C_1

    =~ -\frac{1}{x+1}+C_2

    =~ -(x+1)^{-1}+C_2
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  8. #8
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    Quote Originally Posted by drumist View Post
    Actually both solutions are the same.

    \left(1+\frac{1}{x}\right)^{-1}+C_1

    =~ \frac{1}{1+\frac{1}{x}}+C_1

    =~ \frac{x}{x+1}+C_1

    =~ \frac{x+1-1}{x+1}+C_1

    =~ 1+\frac{-1}{x+1}+C_1

    =~ -\frac{1}{x+1}+C_2

    =~ -(x+1)^{-1}+C_2
    They are the same!!
    Thanks for your help
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