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Math Help - Integrating help

  1. #1
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    Integrating help

    how do you integrate dy/(y(1-y))?

    I got ln|y| + 1/y
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  2. #2
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    Excellent! How did you get that? I would be inclined to use "partial fractions" writing \frac{1}{x(1-x} as \frac{A}{x}+ \frac{B}{1-x}. But integrating that should involve a "ln(1-x)" and I don't see that in your answer.
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  3. #3
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    lol I don't get it. Could you please show me how to work it out and give me the answear
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  4. #4
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    I'll actually give you the full question and hopefully someone can help me.
    dy/dt= y(1-y) and they want us to find the general solution... I know its a seperable equation so I get dy/(y(1-y)) = 1 dt and we're ment to integrate it... though I dont know how. And the answer they give is y(t)= ke^t/(ke^t +1)
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    \frac{1}{x(x+1)}=\frac{x+1-x}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}

    \therefore\int\frac{dx}{x(x+1)}=\int\bigg[\frac{1}{x}-\frac{1}{x+1}\bigg]dx=\ln\left(\frac{x}{x+1}\right)+C
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  6. #6
    Math Engineering Student
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    Quote Originally Posted by Mathstud28 View Post

    \frac{1}{x(x+1)}=\frac{x+1-x}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}

    \therefore\int\frac{dx}{x(x+1)}=\int\bigg[\frac{1}{x}-\frac{1}{x+1}\bigg]dx=\ln\left(\frac{x}{x+1}\right)+C
    ||

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