Question

Find the integral: 1/(y-y^2)dy

Answer 1

I 'm pretty sure you separate into 1/y and 1/(1-y), then using integration by parts, but no matter which variable I choose for u and v, the problem only gets uglier.
\[\int\limits1/(y-y^2)dy\]

Answer 2

I don't think you need to use integration by parts to solve this. Using partial fractions, you can get 2 integrals:
\[\int\limits_{}^{} 1/y\ dy + \int\limits_{}^{} 1/(1-y)\ dy\]
Partial Fractions:
\[1/y(1-y) = A/y + B/(1-y)\]
\[1=A(1-y) + By\]
\[1=(B-A)y+A\]
A & B both equal 1.

So now integrate those 2 integrals and see what you get. I got:
\[\ln \left| y \right| - \ln \left| 1-y \right| +C\]
There is a minus sign before the 2nd ln because when you take the integral of 1/(1-y) and substitute u=(1-y), du=(-1), so you get -ln.

Answer 3

I'm thinking you're correct, but I've got to give it a second for my brain to make sense of it... :-p

Answer 4

I had to do it a few times and reference a couple different sources too; it's the coefficients part of the partial fractions that was tripping me up :P
Try wolframalpha.com if you don't already make use of it :)

Answer 5

That's a huge help, thanks! I've been fighting this problem for about an hour.

Answer 6

Too bad that was only 1 tiny part of the problem I'm dealing with! lol! !&(%^(& differential equations.

Answer 7

I know... I always seem to manage to drop negative signs & powers mysteriously "disappear" because my brain feels like mush. I'm about to review integration techniques for my exam next week: trig substitution and partial fractions :P I'll tackle improper integrals and area/volume problems tomorrow... hopefully, haha. Good luck!

Answer 8

You too, thanks for the help.