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Math Help - Help with indefinite integrals and differntation involing natural logs

  1. #1
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    Help with indefinite integrals and differntation involing natural logs

    Find dy/dx of : Y = ln ( x * (2x + 5)^1/3 ) / 7x^2 + 1


    So to put that in words you have the function y = natural log of x times the cubed root of 2x+5 divided by 7 x squared + 1.

    So far I have it broken down as follows:

    I re wrote it up as ln (x) + ln (2x+5)^1/3 - ln (7x^2 + 1) using properties of logarithms.

    I am having problems now with the - ln (7x^2 +1) part.

    Before I get to far ahead of myself am I even headed in the right direction for what the problem is asking?

    Basically I take finding dy/dx meaning say taking the function y = 3x^2.

    Its derivative would be 6x (dx) and dy/dx would just be 6x.


    Thats my first problem I'm working on.

    I have two more that I have solved and am looking to make sure they are correct.


    1) the indefinite integral of (x^4 / 3 - x^5) dx

    For this one I set U = 3 - X^5 so that makes du = -4x^4.

    So to balance I need to include a -4 in the numerator and subsequently add a factor of -1/4 outside of the integral. which makes it look like

    -1/4 indef integral of (x^4) / (3 - x^5)


    So for this one my final answer looks like -1/4 ln (3 - x^5) + C



    2) the indefinite integral of sec(3x) dx

    u = 3x
    du = 3(dx)
    du/3 = dx

    = 1/3 * indef integral csc(u) du

    = 1/3 * ln [ sec(u) + tan(u) ] + C

    = 1/3 * ln [ sec(3x) + tan(3x) ] + C


    So thats where I'm at, any help would be appreciated !!
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  2. #2
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    y = \ln\left(\frac{x(2x+5)^{\frac{1}{3}}}{7x^2+1}\righ  t)

    y = \ln{x} + \frac{1}{3}\ln(2x+5) - \ln(7x^2+1)

    \frac{dy}{dx} = \frac{1}{x} + \frac{2}{3(2x+5)} - \frac{14x}{7x^2+1}



    \int \frac{x^4}{3-x^5} \, dx

    -\frac{1}{5} \int \frac{-5x^4}{3-x^5} \, dx

    -\frac{1}{5} \ln|3-x^5| + C



    last one is correct, except for the csc(u) you wrote ...
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  3. #3
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    Oh yeah it should be sec(u) which integrates into ln | sec(u) + tan(u) | + C

    So for the first one I was basically down to deriving what I had re arranged and the other two were correct except for me incorrectly deriving 3-x^5. It derives into 5x^4 so to balance you need a -1/5 infront of the intergrand. which makes it -1/5 ln | 3 - x^5 | + C


    Ok I figured i was very close with them and this clears it right up! Thanks!


    Also in the future when I need help how do I post my problems like you have ? It would be easier to have them with the proper symbols and things like how you rewrote my examples in a more formally in math terms.

    Thanks again! Lookin forward to helpin and gettin help when i need it!!
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  4. #4
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    Quote Originally Posted by battleman13 View Post
    Oh yeah it should be sec(u) which integrates into ln | sec(u) + tan(u) | + C

    So for the first one I was basically down to deriving what I had re arranged and the other two were correct except for me incorrectly deriving 3-x^5. It derives into 5x^4 so to balance you need a -1/5 infront of the intergrand. which makes it -1/5 ln | 3 - x^5 | + C


    Ok I figured i was very close with them and this clears it right up! Thanks!


    Also in the future when I need help how do I post my problems like you have ? It would be easier to have them with the proper symbols and things like how you rewrote my examples in a more formally in math terms.

    Thanks again! Lookin forward to helpin and gettin help when i need it!!
    It's done using an inbuilt LaTeX compiler.

    Go to the LaTeX sub-forum and read the tutorials.
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