The partial fraction equation is:
1 /(x(x+1)^2) = A/x + B/(x+1)+C/(x+1)^2
Multiply both sides by x(x+1)^2
1 = A(x+1)^2 + Bx(x+1)+Cx
Eliminate B and C by Letting x = 0:
1 = A(0+1)^2
A = 1
1 = (x+1)^2 + Bx(x+1)+Cx
Eliminate B by letting x = -1
1 = C(-1)
C = -1
1 = (x+1)^2 + Bx(x+1)-x
Let x = 1:
1 = (1+1)^2 + B(1)(1+1)-1
2-4 = 2B
B = -1
The partial fraction expansion is:
1 /(x(x+1)^2) = 1/x -1/(x+1)-1/(x+1)^2
Check:
1/x -1/(x+1)-1/(x+1)^2 = 1/x(x+1)^2/(x+1)^2-1/(x+1)(x(x+1))/(x(x+1)) - 1/(x+1)^2
x/x
1/x -1/(x+1)-1/(x+1)^2 = (x^2+ 2x +1-x^2-x -x)/(x(x+1)^2)
1/x -1/(x+1)-1/(x+1)^2 = 1/(x(x+1)^2)
This checks.
The original integrand is equal to the partial fractions:
int ( (dx) / ( x(x+1)^2 ) ) = int 1/x -1/(x+1)-1/(x+1)^2 dx
Separate into 3 integrals:
int ( (dx) / ( x(x+1)^2 ) ) = int 1/x dx - int 1/(x+1) dx - int 1/(x+1)^2 dx
These integrals are well known:
int ( (dx) / ( x(x+1)^2 ) ) = ln|x| - ln|x+1| + 1/(x+1) + C
