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Derivatives of integrals

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation: dy/dx=(y-y2)/x , for all x[tex]\neq[/tex] 0

    2. Relevant equations

    Integration by Parts: [tex]\int[/tex] u dv = u v - [tex]\int[/tex] v du

    [tex]\int[/tex]lnx= 1/x + C
    [tex]\int[/tex] (1/x) = lnx + C
    dy/dx lnx = 1/x
    dy/dx 1/x = lnx

    3. The attempt at a solution


    [tex]\int[/tex] 1/(y-y2) dy = [tex]\int[/tex] 1/x dx

    [tex]\int[/tex] (1/y)(1/(1-y))dy = lnx + C

    (Integration by parts)
    u=1/x dv=(1/(1-y))dy
    du=lnydy v= -ln(1-y)dy

    -lny / y + [tex]\int[/tex] lny ln(1-y) dy

    And then if I continue and do integration by parts again, it just goes back to the original integral.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 21, 2010 #2


    Staff: Mentor

    Instead of trying to do this by integration by parts, use the technique of partial fractions. In a nutshell you want to rewrite your integral on the left side:
    [tex]\int \frac{dy}{y(1 - y)}[/tex]

    with the fraction broken up into two separate fractions, like so:
    [tex]\int \left[\frac{A}{y} + \frac{B}{1 - y}\right]dy[/tex]

    What you need to do is to find constants A and B so that 1/(y(1 - y)) is identically equal to A/y + B/(1 - y).
  4. Jan 22, 2010 #3
    I don't mean to butt in, but can someone verify that:
    [tex]\frac{d}{dx}(\frac{1}{x}) = \ln{x}?[/tex]
    Also, the integral:
    [tex]\int{\ln{x}} \left \left dx = \frac{1}{x}?[/tex]
    I'm confused...
    Many thanks.
  5. Jan 22, 2010 #4
    I think you've confused the concept of integral with that of derivative as

    [tex]\frac{d}{dx}(\ln(x)) = \frac{1}{x}[/tex]
    [tex]\int{\frac{1}{x}} \left \left dx = \ln(x)+C[/tex].

    And yeah both are correct.

  6. Jan 22, 2010 #5


    User Avatar
    Science Advisor

    No, no one can verify that- it's not true. What is true is that
    [tex]\frac{d}{dx}(\frac{1}{x})= -\frac{1}{x^2}[/tex]
    because [itex]\frac{1}{x}= x^{-1}[/itex] and the derivative of [itex]x^n[/itex] is [itex]nx^{n-1}[/itex].

    No, that's also not true. [itex]\int ln(x) dx= ln(x)(x- 1)+ C[/itex]

    You have these both backwards: [itex]d(ln(x))/dx= 1/x[/itex] and [itex]\int 1/x dx= ln(x)+ C[/itex].
  7. Jan 22, 2010 #6
    Thanks both Altabeh and Halls.
    I didn't consider them to be true, hence why I said I was confused. It was from the OP's relevant equations section.
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