How do I find ∫dy /(y^2 - y)

Use partial fraction decomposition. First write
1 / (y^2 - y) = A / (y-1) + B / y

Then solve for A and B to get A = 1, B = -1. The question then becomes how to solve
∫ [ 1 / (y-1) - 1 / y ] dy,
which you should know how to finish.

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== My original problem states "Prove that y = 1/(Ce^2 +1) is a solution of y'' + y^2 = y'

First, you need to correct the equation
y = 1/(Ce^2 +1),
which seems to be missing any mention of an independent variable (t, x, etc.?)

Then just find y' and y" by differentiation.

Finally, just substitute these values into the differential equation and show that the resulting equation is true.

Decompose this expression into partial fractions:
1 / (y² - y) = 1 / [y(y - 1)] = A / y + B / (y - 1)
1 = A(y - 1) + By
Let y = 0,
-A = 1
A = -1
Let y = 1,
B = 1
1 / (y² - y) = -1 / y + 1 / (y - 1)
1 / (y² - y) = 1 / (y - 1) - 1 / y

Integrate the expression term by term using this result:
∫ 1 / (y² - y) dy = ∫ [1 / (y - 1) - 1 / y] dy
∫ 1 / (y² - y) dy = ∫ 1 / (y - 1) dy - ∫ 1 / y dy
∫ 1 / (y² - y) dy = ln|y - 1| - ln|y| + C

∫dy /(y^2 - y)=∫dy /y*(y - 1)

1/y*(y-1)=[1/(y-1)] -[1/y]

=>∫dy /(y^2 - y)=∫dy /(y-1) -∫dy /y

=lnI y-1 I -ln IyI +c

=ln l 1-(1/y) l+c

1 / (y^2 - y)

= 1/ y(y - 1)

split into partial fractions

= 1/(y-1) - 1/y

==> ∫dy /(y^2 - y) = ∫dy /(y - 1) -∫dy /y

= ln I y - 1 I - ln I y I + C