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Math Help - Please help to solve this Integration question

  1. #1
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    Please help to solve this Integration question

    Hi Guys,

    Please help to solve this question. I do not know how to solve this.
    Your help is greatly appriecated.
    The answer is 2.

    Thank you
    Attached Thumbnails Attached Thumbnails Please help to solve this Integration question-math.jpg  
    Last edited by aeonboy; November 19th 2007 at 01:31 AM.
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  2. #2
    Senior Member Peritus's Avatar
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    use the substitution x = e^t and then perform integration by parts.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aeonboy View Post
    Hi Guys,

    Please help to solve this question. I do not know how to solve this.
    Your help is greatly appriecated.
    The answer is 2.

    Thank you
    I like Peritus' substitution

    alternatively:

    we could split the integral up and find,

    \int_1^e \frac {\ln x}{\sqrt{x}}~dx + \int_1^e \frac 1{\sqrt{x}}~dx

    the last integral can be done using the power rule, for the first, i see two ways of approaching it.

    the easier of the two ways is to make the substitution u = \sqrt{x}, and continue by integration by parts, which will be easy (you may even know the integral of ln x by heart, so integration by parts won't be necessary)

    the second way is to multiply the first integral by \frac {\sqrt{x}}{\sqrt{x}} and then use the substitution u = \ln x. again you would continue by parts, but this one would be a little harder i think
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  4. #4
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    Quote Originally Posted by Peritus View Post
    use the substitution x = e^t and then perform integration by parts.
    Hi Peritus,

    Thanks for your help but i am not sure how to use the substitution and then perform integration by parts.

    Could anyone guide me with the steps?

    Thank you
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by aeonboy View Post
    Hi Peritus,

    Thanks for your help but i am not sure how to use the substitution and then perform integration by parts.

    Could anyone guide me with the steps?

    Thank you
    using Jhevon's solution, we are left to find the \int \frac{ln x}{\sqrt x} \, dx

    if we set, u = \sqrt x \implies u^2 = x (we can assume this is true since x is positive, otherwise, we need the absolute value of x..)

    now, du = \frac{1}{2\sqrt x} \, dx \implies 2du = \frac{1}{\sqrt x} \, dx

    so we have \int {2 ln \, u^2 \, du} = 4\int {ln \, u \, du}

    and like what he said, you should know the integral of ln x by heart..
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  6. #6
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    Consider f(x)=x\ln x. Take its derivative f'(x)=\ln x+1.

    Integrate f(x)+k=\int\ln x\,dx+x.

    And we happily get \int\ln x\,dx=x\ln x-x+k.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Consider f(x)=x\ln x. Take its derivative f'(x)=\ln x+1.

    Integrate f(x)+k=\int\ln x\,dx+x.

    And we happily get \int\ln x\,dx=x\ln x-x+k.
    hehe. i like this approach, when you use a derivative to find the integral. it is a neat concept. sometimes it is hard to come up with the function to differentiate, but when you can come up with it, it seems all too obvious. (in fact, the method of integration by parts was developed using this approach. what they did was to undo the product rule)

    for the benefit of the class, i shall show how we find the integral of ln(x) using the by parts method.

    \int ln x~dx

    the trick here is to note that \ln x = 1 \cdot \ln x, obvious enough

    Thus, if we take u = \ln x \implies u' = \frac 1x and v' = 1 \implies v = x, we have:

    \int uv' = uv - \int v'u

    \Rightarrow \int \ln x = x \ln x - \int x \cdot \frac 1x~dx

    = x \ln x - \int ~dx

    = x \ln x - x + C


    An alternative method to find the integral of \ln x is the following. (i actually just thought of this now, i've never seen this done before, but the technique is a very familiar one)

    \int \ln x~dx = \int \frac xx \ln x~dx

    By substitution, let u = \ln x \implies x = e^u

    \Rightarrow du = \frac 1x~dx

    Thus, our integral becomes

    \int u e^u~du = ue^u - \int e^u~du (by parts)

    = u e^u - e^u + C

    back-substitute:

    = \ln x \cdot e^{\ln x} - e^{\ln x} + C

    = x \ln x - x + C

    as desired


    of course, this way is a bit more work than the last, but it's nice to know several ways to skin a cat...if skinning a cat is your thing
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  8. #8
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    I have managed to solve it already, thanks to all of you for your help
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