The partial fraction equation is:
1 /(x(x+1)^2) = A/x + B/(x+1)+C/(x+1)^21x(x+1)2=Ax+Bx+1+C(x+1)2
Multiply both sides by x(x+1)^2x(x+1)2
1 = A(x+1)^2 + Bx(x+1)+Cx1=A(x+1)2+Bx(x+1)+Cx
Eliminate B and C by Letting x = 0x=0:
1 = A(0+1)^21=A(0+1)2
A = 1A=1
1 = (x+1)^2 + Bx(x+1)+Cx1=(x+1)2+Bx(x+1)+Cx
Eliminate B by letting x = -1x=−1
1 = C(-1)1=C(−1)
C = -1C=−1
1 = (x+1)^2 + Bx(x+1)-x1=(x+1)2+Bx(x+1)−x
Let x = 1x=1:
1 = (1+1)^2 + B(1)(1+1)-11=(1+1)2+B(1)(1+1)−1
2-4 = 2B2−4=2B
B = -1B=−1
The partial fraction expansion is:
1 /(x(x+1)^2) = 1/x -1/(x+1)-1/(x+1)^21x(x+1)2=1x−1x+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/x1x−1x+1−1(x+1)2=1x(x+1)2(x+1)2−1x+1x(x+1)x(x+1)−1(x+1)2xx
1/x -1/(x+1)-1/(x+1)^2 = (x^2+ 2x +1-x^2-x -x)/(x(x+1)^2)1x−1x+1−1(x+1)2=x2+2x+1−x2−x−xx(x+1)2
1/x -1/(x+1)-1/(x+1)^2 = 1/(x(x+1)^2)1x−1x+1−1(x+1)2=1x(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∫(dxx(x+1)2)=∫1x−1x+1−1(x+1)2dx
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∫(dxx(x+1)2)=∫1xdx−∫1x+1dx−∫1(x+1)2dx
These integrals are well known:
int ( (dx) / ( x(x+1)^2 ) ) = ln|x| - ln|x+1| + 1/(x+1) + C∫(dxx(x+1)2)=ln|x|−ln|x+1|+1x+1+C
