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Math Help - Integration

  1. #1
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    Integration

    I did this exercise but the book answer was =x-e^-x+c. And mine was
    = -x-e^-x+c. Thanks for ur help.
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  2. #2
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    Quote Originally Posted by ron007 View Post
    I did this exercise but the book answer was =x-e^-x+c. And mine was
    = -x-e^-x+c. Thanks for ur help.

    \int \frac{e^x+1}{e^x}dx=\int (1+e^{-x}) dx =x - e^{-x}+Const

    As a check you can try differentiation this and your solution

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    \int \frac{e^x+1}{e^x}dx=\int (1+e^{-x}) dx =x - e^{-x}+Const

    As a check you can try differentiation this and your solution

    RonL
    Thanks. But can u show the process to see my error.
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    \int \frac{e^x+1}{e^x}dx=\int (1+e^{-x}) dx =x - e^{-x}+Const

    As a check you can try differentiation this and your solution

    RonL
    Quote Originally Posted by ron007 View Post
    Thanks. But can u show the process to see my error.
    There is virtualy nothing to show:

    ] \int \frac{e^x+1}{e^x}dx

    divide through by the e^x in the denominator to get:

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

    Distribute the integral over the addition:

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

    But the two integrals on the right are both standard integrals:

    \int 1\ dx=x+C

    \int e^{-x} dx = -e^{-x}+K

    so:

    \int \frac{e^x+1}{e^x}dx=\int (1+e^{-x}) dx = \int 1\ dx + \int e^{-x} dx=x-e^{-x}+Const

    RonL
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    Thanks...
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