Integration, how to solve integration of 1/(y(1-y))?

Question

1/(y(1-y))

it should be pretty basic but I am lost. help!

Answer 1

Let 1 / (y (1 - y) = A / y + B / (1- y)

1 = A(1 - y) + By

When y = 1, B = 1

When Y = 0, A = 1

I = ∫ 1 / y dy + ∫1 / (1 - y) dy

I = log y - log(1 - y) + c

I = K log {y / (1 - y) }

Answer 2

1/y(1-y) can be written in the form of A/y + B/(1-y). interegate them seperately. here A and B are constants only.

Answer 3

In case you want to look this up in your textbook, the appropriate section would be called "Partial Fraction Decomposition". Good luck.

Answer 4

1/(y(1-y)) = A/y + B/(1 - y) =
1/y + 1/(1 - y)
∫dy/y - ∫dy/(y - 1) =
Ln(y) - Ln(y - 1) + Ln(C) =
Ln(Cy/(y - 1))

Answer 5

y-y2(2 is to the square)