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Math Help - Stuck in loop using sub method for integrals, work scanned

  1. #1
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    Stuck in loop using sub method for integrals, work scanned

    I'm doing so much math I think I got a blank on the substitution method for solving integrals.

    My work is scanned in the attachment. I think I'm on the right track but there is something I'm forgetting because I'm stuck in a loop.
    Attached Thumbnails Attached Thumbnails Stuck in loop using sub method for integrals, work scanned-loopingsub.png  
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    MHF Contributor harish21's Avatar
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     \int \frac{30x^2}{x^{3}-6} dx

    let  u = x^{3}-6

    then  du = 3x^2 dx \implies dx = \frac{du}{3x^2}

    so you have:

     \int \frac{30x^2}{x^{3}-6} dx= \int \frac{30x^2}{3x^2} \frac{1}{u}du = 10[log(u)]+C =  10[log(x^3-6)]+C
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    \displaystyle 5 \cdot \frac{x^5}{x^3-6} =

    \displaystyle 5 \left( x^2 + \frac{6x^2}{x^3-6}\right) =

    \displaystyle 5x^2 + 10 \cdot \frac{3x^2}{x^3-6}


    \displaystyle 5 \int x^2 \, dx + 10\int  \frac{3x^2}{x^3-6} \, dx

    \displaystyle \frac{5x^3}{3} + 10\ln|x^3-6| + C
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    I can follow these steps and understand them except the part where we get to the last line and we introduce ln. I know 1/u = ln|u| but where does the numerator 3x^2dx go?

    Edit: Also, the integral has limits of integration 3 and 4, what do we do with this? Normally when I see limits like this I apply a formula F(b) - F(a), would this be the step after the final step shown above?
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    MHF Contributor harish21's Avatar
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    After supposing  u = x^3-6, and finding du = 3x^2 dx you get:

    \int \frac{30x^2}{x^{3}-6} dx= \int \frac{30x^2}{3x^2} \frac{1}{u}du

    you should be able to see that the x^2 in the numerator and denominator cancel each other other out, and 30/3 = 10...

    you book should have examples on how to use the limits of integration.


     \displaystyle \left[\frac{5x^3}{3} + 10\ln|x^3-6|\right]^4_3
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  6. #6
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    Don't forget (lnf(x))'=F'(x)/F(x).Here:
    (3x^2)/(x^3-6)=
    (x^3-6)'/(x^3-6)
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  7. #7
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by solidstatemath View Post
    I can follow these steps and understand them except the part where we get to the last line and we introduce ln. I know 1/u = ln|u| but where does the numerator 3x^2dx go?

    Edit: Also, the integral has limits of integration 3 and 4, what do we do with this? Normally when I see limits like this I apply a formula F(b) - F(a), would this be the step after the final step shown above?
    The integral has been done so I won't bother with that. For the limits you may also leave it in terms of u and change the limits using the original subsitution:

    u(3) = 3^3-6 = 21 and u(4) = 4^3-6 = 58


    \left[\: 10 \ln |u| \: \right]^{58}_{21}
    Last edited by e^(i*pi); October 9th 2010 at 12:53 PM. Reason: copied incorrect integral
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  8. #8
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    Lol yes, the 10 is what is left after simplification... I even did that myself when I was stuck in my loop. :P Thanks.
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    By using limits of integration I apply F(4) - F(3) and this is what I get:

    10ln58 - 10ln21 = ~10.16 and it's wrong.

    What could I be doing wrong?
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  10. #10
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    Quote Originally Posted by solidstatemath View Post
    By using limits of integration I apply F(4) - F(3) and this is what I get:

    10ln58 - 10ln21 = ~10.16 and it's wrong.

    What could I be doing wrong?
    you forgot the term \displaystyle \frac{5x^3}{3} in the antiderivative
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