Recent content by Europio2

¿Do you mean that? \begin{equation} y = -\int -1/(1-x)^2 \cdot dx = -\int -(1-x)^{-2} \cdot dx = -\frac{(1-x)^{-1}}{-1} = \frac{1}{1-x} \end{equation} But if I change the exponent to 5, as in the other post, and I do the same, I don't get the correct one. \begin{equation} y = -\int...

I do this -4*(1-x)^-4 = 4*(x-1)^-4, that's why I change the sign.

But that formula is valid for any other exponent I think. Look at this (this is correct). \begin{equation} y = \int 1/(1-x)^5 \cdot dx = \int (1-x)^{-5} \cdot dx = \frac{(1-x)^{-4}}{-4} = \frac{1}{4(x-1)^4} \end{equation} That's why I am confused.

Sorry. This is what I do. ∫1/(1-x)^2dx = ∫(1-x)^-2dx = (1-x)^-1/-1 = 1/(x-1)

Homework Statement I know its easy, but I'm making a mistake somewhere that is making me crazy. I want to solve \begin{equation} y = \int 1/(1-x)^2 \cdot dx \end{equation} I use de sixth formula in this PDF, but it does not work...