1. ## Integrating help

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

I got ln|y| + 1/y

2. Excellent! How did you get that? I would be inclined to use "partial fractions" writing $\displaystyle \frac{1}{x(1-x}$ as $\displaystyle \frac{A}{x}+ \frac{B}{1-x}$. But integrating that should involve a "ln(1-x)" and I don't see that in your answer.

3. lol I don't get it. Could you please show me how to work it out and give me the answear

4. 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)

5. $\displaystyle \frac{1}{x(x+1)}=\frac{x+1-x}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$

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

6. Originally Posted by Mathstud28

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

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