integrate

**Rui** · May 10th 2008, 05:04 AM

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!

**galactus** · May 10th 2008, 05:10 AM

You didn't finish factoring.

$\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}$

Now, it's easier, huh?.

$\lim_{x\rightarrow{1}}\frac{x-1}{(x-1)(x+1)(x+2)}=\frac{1}{6}$

$\lim_{x\rightarrow{-1}}\frac{x+1}{(x-1)(x+1)(x+2)}=\frac{-1}{2}$

$\lim_{x\rightarrow{-2}}\frac{x+2}{(x-1)(x+1)(x+2)}=\frac{1}{3}$

$\frac{1}{6(x-1)}-\frac{1}{2(x+1)}+\frac{1}{3(x+2)}$

Now, integrate.

**Moo** · May 10th 2008, 05:13 AM

Hello,

i know have to use partial fractions.
A/ (x^2-1) + B /(x+2) = 1/ (x^3 + 2x^2- x-2)

False

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 $(x^3+2x^2-x-2)=(x-1)(x+1)(x+2)$

Thus you have to find A, B, C such that :

$\frac{1}{x^3+2x^2-x-2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}$

A/ (x^2-1)

More generally, if you have a term of degree 2 in the denominator, you should have a polynomial of degree 1 in the numerator.

So here, it should have been $\frac{Ax+A'}{x^2-1}$

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