You didn't finish factoring.
Now, it's easier, huh?.
Now, integrate.
integrate 1/(x^3 + 2x^2- x-2) dx
i know have to use partial fractions.
A/ (x^2-1) + B /(x+2) = 1/ (x^3 + 2x^2- x-2)
and found B =1/3 and A =1/3 or 1 (don't know which one to use so i used A= 1/3)
after integrate i got ln(3x^2-3) + ln(3x+6) not sure if it is right..but i have a feeling that it's wrong. Help!
Hello,
Falsei know have to use partial fractions.
A/ (x^2-1) + B /(x+2) = 1/ (x^3 + 2x^2- x-2)
Ok (x^3+2x^2-x-2)=(x^2-1)(x+2), but you have to continue. (x^2-1)=(x-1)(x+1)
So
Thus you have to find A, B, C such that :
More generally, if you have a term of degree 2 in the denominator, you should have a polynomial of degree 1 in the numerator.A/ (x^2-1)
So here, it should have been