Question

integrate 1/(x^2-x)dx

Answer 1

Partial fraction decomposition.

Answer 2

just kidding [: you can factor out an x and get 1/[x(x-1)], and then get two integrals, 1/xdx and 1/(x-1)dx

Answer 3

I have A/x + B/(x-1) but i dont know how to solve it from there. I missed class the day the teacher taught it :(

Answer 4

Careful with splitting up integrals,
\[\Large \int\frac{1}{x(x-1)}dx \]
It's not working to split this integral up and evaluate it individually.

Answer 5

\[\Large \frac{1}{x(x-1)}= \frac{A}{x}+ \frac{B}{x-1} \]
So you get.

\[\Large 1=A(x-1)+Bx = (A+B)x-A \]
Matching the coefficients:
\[\Large A+B=0 \\
\Large-A=1 \]

Answer 6

lol i realized that, sorry :/ but partial fractions can be solved like this:
\[1/(x^2-x)=A/x+B/(x-1)\]
now, you want to figure out what A and B are, so the first step would be to cancel out the denominators. the easiest way to do this is multiply each part by (x^2-x)/(x^2-x). This will give you the equation \[1=A(x-1) + B(x)\]
Now, to solve for A, you want B to equal zero, so you set x=0, since B*0=0. This will give you the equation 1=A*-1, and thus, A = -1.
similarly, to solve for B, you set x=1 so you can get A to equal zero. This gives you the equation 1=A(1-1) + B(1), or simply 1=B.
you now have two integrals:

Answer 7

\[\int-1/xdx + \int1/(x-1)dx\]

Answer 8

and those integrals should be fairly easier to solve, yeah? :)

Answer 9

Yes! Thank you!!!

Answer 10

SO i get -ln(x) + ln(x-1) + C ,right?

Answer 11

yep :)