How do you find the indefinite integral of #int (x(x+2))/(x^3+3x^2-4)#?

1 Answer
Jun 1, 2018

# 1/3ln|(x+2)^2(x-1)|+C, OR, ln|(x+2)^(2/3)(x-1)^(1/3)|+C#.

Explanation:

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.!