Here's the question you clicked on:
whatisthequestion
Evaluate the integral. The integral will be entered below. The title of this section is Intergration of Rational Fractions by Partial Fractions
\[\int\limits_{}^{}\frac{ x^2+2x-1 }{ x^3-x }dx\]
I can simplify it down to \[\int\limits_{}^{}\frac{ A }{ x }+\frac{ B }{ x-1 }+\frac{ C }{ x+1 }\] but get lost when solving for A, B and C
oops forgot a dx there
do u want answer or explanation
explanation my math teacher was very confusing when going over this section (thick accent)
I probably screwed up because its three terms now instead of two
I had \[x^2+2x-1=Ax+B(x-1)+C(x+1)\] which is obviously wrong
yea this method is past l'Hopital's rule which I think you said you did not know earlier so it might be a little higher than your current level of math course
first factor the function
which I did getting x(x-1)(x+1) on the bottom
yes that is true
\[\large \frac{x^2+2x-1}{x(x-1)(x+1)} \qquad =\qquad \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\]You should set it up like this. It looks like you're on the right track. You just didn't multiply through correctly.
Multiply both sides by the denominator on the left. :O
O GOD top ten list of signs I need to go to bed
so the correct answer is \[x^2+2x-1=A(x+1)(x-1)+Bx(x+1)+Cx(x-1)\]\[=(A+B+C)X^2+(B-C)X-A\]\[A=1, B=1, C=-1\]
Yah that looks right :)