Integrate using partial fractions, answer discrepancy My attempt:
$\eqalign{
  & \int {{{{x^2} + x + 2} \over {3 - 2x - {x^2}}}} dx  \cr 
  & {{ - 1( - {x^2} - 2x + 3) - 2x + 3 + x + 2} \over {3 - 2x - {x^2}}}  \cr 
  &  =  - 1 + {{ - x + 5} \over { - {x^2} - 2x + 3}}  \cr 
  &  =  - 1 + {{5 - x} \over { - {x^2} - 2x + 3}}  \cr 
  &  =  - 1 + {{x - 5} \over {{x^2} + 2x - 3}} \cr} $
I think i've messed up somewhere here, when i've tried to manipulate the minus signs. Carrying on:
Finding ${{x - 5} \over {{x^2} + 2x - 3}}$ as a partial fraction:
$\eqalign{
  & {{x - 5} \over {{x^2} + 2x - 3}} \equiv {A \over {(x + 3)}} + {B \over {(x - 1)}}  \cr 
  & x - 5 = A(x - 1) + B(x + 3)  \cr 
  & x = 1:  \cr 
  &  - 4 = 4B  \cr 
  & B =  - 1  \cr 
  & x =  - 3:  \cr 
  &  - 8 =  - 4A  \cr 
  & A = 2  \cr 
  & so:  \cr 
  & {{{x^2} + x + 2} \over {3 - 2x - {x^2}}} \equiv  - 1 + {2 \over {(x + 3)}} - {1 \over {(x - 1)}}  \cr 
  & \int { - 1 + {2 \over {(x + 3)}} - {1 \over {(x - 1)}}dx}   \cr 
  &  =  - x + \ln |{(x + 3)^2}| - \ln |(x - 1)| + C  \cr 
  &  =  - x + \ln \left( {\left|{{{{(x + 3)}^2}} \over {(x - 1)}}\right|} \right) + C \cr} $
However this is the wrong answer, the right answer is:
$$\int {f(x)dx}  =  - x + \ln \left( {\left|{{{{(3 + x)}^2}} \over {(1 - x)}}\right|} \right) + C$$
Thank you for your help in advance.

No worries!
$$- x + \ln \left( {\left|{{{{(x + 3)}^2}} \over {(x - 1)}}\right|} \right) + C =  - x + \ln \left( {\left|{{{{(3 + x)}^2}} \over {-(1 - x)}}\right|} \right) + C = - x + \ln \left( {\left|{{{{(3 + x)}^2}} \over {(1 - x)}}\right|} \right) + C = $$
