Let, #I=int(x(x+2))/(x^3+3x^2-4)dx#.
The sum of the co-effs. of the poly. in the Dr. is #0#.
#:. (x-1)# is its factor.
We have, #x^3+3x^2-4=ul(x^3-x^2)+ul(4x^2-4x)+ul(4x-4)#,
#=x^2(x-1)+4x(x-1)+4(x-1)#,
#=(x-1)(x^2+4x+4)#,
#=(x-1)(x+2)^2#.
#:. (x(x+2))/(x^3+3x^2-4)=(x(x+2))/((x-1)(x+2)^2)=x/((x-1)(x+2))#.
Observe that, #2(x-1)+1(x+2)=3x#. Therefore,
#I=int(x(x+2))/(x^3+3x^2-4)dx#
#=intx/((x-1)(x+2))dx#,
#=1/3int(3x)/((x-1)(x+2))dx#,
#=1/3int{2(x-1)+1(x+2)}/((x-1)(x+2))dx#,
#=1/3int{(2(x-1))/((x-1)(x+2))+(1(x+2))/((x-1)(x+2))}dx#,
#=1/3int{2/(x+2)+1/(x-1)}dx#,
#=1/3{2ln|(x+2)|+ln|(x-1)|}#.
#:.I=1/3ln|(x+2)^2(x-1)|+C, or, #
# I=ln|(x+2)^(2/3)(x-1)^(1/3)|+C#.
Enjoy Maths.!
