How do you integrate #int(x(x+2))/(x^3 +3x^2 -4) dx#?

Let us begin by factorizing the denominator. It is easy to see that #x^3+3x^2-4# vanishes when #x=1#, so that #(x-1)# is a factor.

#x^3+3x^2-4 = x^3-x^2+4x^2-4x+4x-4 #
# = x^2(x-1)+4x(x-1)+4(x-1) = (x-1)(x^2+4x+4) = (x-1)(x+2)^2#

Thus, the integrand simplifies considerably to

#{x(x+2)}/{x^3+3x^2-4} = {x(x+2)}/{(x-1)(x+2)^2} = x/{(x-1)(x+2)}#

We try the partial fraction expression

#x/{(x-1)(x+2)} = A/{x-1}+B/{x+2}#

Thus
#x = A(x+2) +B(x-1)#

substituting #x=1# and #x=-2#, respectively, gives

#1 = A times 3 implies A =1/3, -2 = B times (-3) implies B=2/3#

Thus

#x/{(x-1)(x+2)} = 1/3 1/{x-1}+2/3 1/{x+2}#

Finally

#int {x(x+2)}/{x^3+3x^2-4} dx = int ( 1/3 1/{x-1}+2/3 1/{x+2}) dx#
#qquad = 1/3 ln|x-1|+ 2/3 ln|x+2|+C#

# "We want to find:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx. #

# "Taking a quick second look, multiplying out the numerator," #
# "we see:" #

# \qquad \qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx \ = \ int \ { x^2 + 2 x}/{ x^3 + 3 x^2 - 4 } \ dx. #

# "Observing the derivative of the denominator:" \qquad 3 x^2 + 6 x, #
# "which just happens to be proportional to numerator:" \ \ x^2 + 2 x, #
# "[this situation is a rare accident -- but welcome], we can finish" #
# "directly as:" #

# \qquad \qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx \ = \ int \ { x^2 + 2 x }/{ x^3 + 3 x^2 - 4 } \ dx. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ 1/3 int \ { 3 ( x^2 + 2 x ) }/{ x^3 + 3 x^2 - 4 } \ dx. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ 1/3 int \ { 3 x^2 + 6 x }/{ x^3 + 3 x^2 - 4 } \ dx. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ 1/3 int \ { ( x^3 + 3 x^2 - 4 )' }/{ x^3 + 3 x^2 - 4 } \ dx. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ 1/3 ln | x^3 + 3 x^2 - 4 | + C. #

# "Thus:" #

# \qquad \qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx \ = \ 1/3 ln | x^3 + 3 x^2 - 4 | + C. #