# Integrating help

• Oct 28th 2008, 03:57 PM
Unt0t
Integrating help
how do you integrate dy/(y(1-y))?

I got ln|y| + 1/y
• Oct 28th 2008, 04:07 PM
HallsofIvy
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.
• Oct 28th 2008, 04:16 PM
Unt0t
lol I don't get it. Could you please show me how to work it out and give me the answear
• Oct 28th 2008, 04:37 PM
Unt0t
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)
• Oct 28th 2008, 04:57 PM
Mathstud28
$\frac{1}{x(x+1)}=\frac{x+1-x}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$

$\therefore\int\frac{dx}{x(x+1)}=\int\bigg[\frac{1}{x}-\frac{1}{x+1}\bigg]dx=\ln\left(\frac{x}{x+1}\right)+C$
• Oct 28th 2008, 05:13 PM
Krizalid
Quote:

Originally Posted by Mathstud28

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

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

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:)