Question

How do I use partial fractions to find the integral of 1/(x^2 - x) dx?

Answer 1

I have the integral of 1/ (x-1)(x) = A/(x-1) + B/(x)
So
the integral of 1 = A(x) + B(x+1)? I don't know if that's right, and I don't know what to do next.

Answer 2

you have
\[ \frac{1}{x(x-1)}= \frac{A}{(x-1)}+ \frac{B}{x} \]
or
\[ \frac{1}{x(x-1)}= \frac{Ax}{x(x-1) }+\frac{B(x-1)}{x(x-1)}\]
or
\[ \frac{1}{x(x-1)}= \frac{Ax+B(x-1)}{x(x-1) }\]

looking at this, we see that the tops of both sides must be equal
\[ 1= Ax+B(x-1) \]
or
\[ (A+B)x -B = 1 \]
or
\[ (A+B)x -B = 0x+ 1 \]

now match corresponding coefficients to find 2 equations.
in other words, the coefficient on the x terms must be equal:
A+B= 0
and the constant terms must be equal
-B= 1

now solve for A and B. use them in your original fractions
which can be integrated (you get a natural log)

Answer 3

Cool! I was just missing the rearranging step from (x-1)B+Ax=1 to (A+B)x-B=1. Now it makes sense. Thanks! :D