`(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)`

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