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Homework Help: Help with log integration/differntiation

  1. Aug 14, 2008 #1
    I'm having trouble on these two problems. Can anyone give me a step by step explanation on how to do them (or one of them)?

    7. Find f'(x), if
    f(x) = [tex]\left|\frac {x^2\left((3x + 2)^{\frac {1}{3}}\right)}{(2x - 3)^3}\right|[/tex]

    13. Perform the integration:
    (a) [tex]\int(x-1)e^{-x^2+2x}dx[/tex]
    (b) [tex]\int\frac{1}{x}e^{-2\log_3(x)}dx[/tex]
    Last edited: Aug 14, 2008
  2. jcsd
  3. Aug 14, 2008 #2


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    This has nothing to do with a logarithm! Do it as two separate problems:
    [tex]f(x)= x^2(3x+2)^{1/3}(2x-3)^{-3}[/tex] if [tex]x\ge 0[/tex]
    [tex]f(x)= -x^2(3x+2)^{1/3}(2x-3)^{-3}[/tex] if x< 0
    Use the chain rule and product rule

    The derivative of -x2+ 2x is -2x+2= -2(x-1). Use the substitution u= x2-2x.

    Finally a problem involving a lograrithm!

    -2 log3(x)= log3(x3)= -2ln(x)/ln(3) so
    [tex]e^{-2 log_3(x)}= 3^ln(x)[/tex]. Let u= ln(x).
  4. Aug 14, 2008 #3


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    ??? If g(x)=x^2*(3x+2)^(1/3)*(2x-3)^(-3), (so f(x)=|g(x)|), you need to find the regions where g(x) is positive or negative. Not where x is positive or negative. Find the x values where g(x) is 0 or undefined. Those are the endpoints of the intervals where g(x) has uniform sign.
  5. Aug 14, 2008 #4
    Thanks for the help so far,

    trying 13 i get
    a. u=-x^2 + 2x
    du = -2(x-1)dx
    int((x-1)/(-2(x-1)) e^u) su
    int( -1/2 e^u du) = -1/2 e^u u' + c = (x-1)(e^(-x^2+2x)) + c

    so, I have int(1/x e^(-2lnx/ln3)) dx
    du = 1/x dx
    int (e^(-2u/ln3)) du = -2/ln3 e^(-2u/ln3) + c = -2/ln3 e^(-2lnx/ln3) + c = -2/ln3 e^(-2 log_3 x) + c

    is this correct?
    i'm not quite sure where i'd put the absolute value signs, but i know i need them.
  6. Aug 14, 2008 #5


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  7. Aug 15, 2008 #6
    Based on Halls of Ivy / Dick 's help, I get this for 1

    [tex]\frac{-(x)(12x^2-26x-27)}{(2x-3)^{-4}(3x+2)^{-2/3}}[/tex] if [tex]{x} > \frac{3}{2}[/tex]or [tex]{x} \leq \frac{-2}{3}[/tex]
    [tex]\frac{-(x)(12x^2-26x-27)}{(2x-3)^{-4}(3x+2)^{-2/3}}[/tex] if [tex] \frac{3}{2} < {x} < \frac{-2}{3}[/tex]

    is this right?
    Last edited: Aug 15, 2008
  8. Aug 15, 2008 #7


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    No... There should be a sign difference between the two expressions on different intervals. And that derivative doesn't look right. The powers of the factors in the denominator should be positive. And I don't get the same quadratic you have in the numerator.
  9. Aug 15, 2008 #8
    Oh, sorry.
    I meant

    [tex]\frac{-(x)(12x^2-26x-27)}{(2x-3)^{4}(3x+2)^{2/3}}[/tex] if [tex]{x} > \frac{3}{2}[/tex]or [tex]{x} \leq \frac{-2}{3}[/tex]
    [tex]\frac{(x)(12x^2-26x-27)}{(2x-3)^{4}(3x+2)^{2/3}}[/tex] if [tex] \frac{3}{2} < {x} < \frac{-2}{3}[/tex]
  10. Aug 15, 2008 #9


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    That's better. But I still don't get the same coefficients you do on the quadratic part.
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