Precalculus, Chapter 7, 7.4, Section 7.4, Problem 53

(4x^2-1)/(2x(x+1)^2)
(4x^2-1)/(2x(x+1)^2)=A/(2x)+B/(x+1)+C/(x+1)^2
(4x^2-1)/(2x(x+1)^2)=(A(x+1)^2+B(2x)(x+1)+C(2x))/(2x(x+1)^2)
:.(4x^2-1)=A(x+1)^2+B(2x)(x+1)+C(2x)
4x^2-1=A(x^2+2x+1)+B(2x^2+2x)+2Cx
4x^2-1=Ax^2+2Ax+A+2Bx^2+2Bx+2Cx
4x^2-1=(A+2B)x^2+(2A+2B+2C)x+A
Therefore from the above,
A+2B=4
2A+2B+2C=0
A=-1
Solve the above equations for getting the values of A, B and C,
Substitute back the value of A in equation 1,
-1+2B=4
2B=4+1=5
B=5/2
Substitute back the values of A and B in equation 2,
2(-1)+2(5/2)+2C=0
-2+5+2C=0
2C+3=0
2C=-3
C=-3/2
:.(4x^2-1)/(2x(x+1)^2)=-1/(2x)+5/(2(x+1))-3/(2(x+1)^2)
(4x^2-1)/(2x(x+1)^2)=(1/2)(-1/x+5/(x+1)-3/(x+1)^2)