Integrate the following w.r.t.x.
`(3x+1)/((x-2)^(2)(x+2))`

Explanation

`"Let I=int(3x+1)/((x-2)^(2)(x+2))dx`
`"Let "(3x+1)/((x-2)^(2)+(x+2))dx`
`therefore 3x+1=A(x-2)(x+2)+B(x+2)+C(x-2)^(2)`
`"Put"x-2=0,i.e, x=2"we get "`
`"Put x-2=0 i.e,x=2 we get"`
`3(2)+1=A(0)(4)+B(4)+C(0)`
`therefore 7=AB" "therefore (7)/(4)`
`"Put x+2=0, i.e. x=-2 we get "`
`3(-2)+1=A(-4)(0)+B(0)+C(-4)^(2)`
`therefore -5=16C" "therefore C=-(5)/(16)`
Put x=0, we get,
`3(0)+1=A(-2)(2)+B(2)+C(-2)^(2)`
`therefore 1=-4A+(7)/(2)-(5)/(4)`
`therefore 1=-4A+(7)/(2)-(5)/(4)`
`therefore 4A=(9)/(4)-1=(5)/4 " "A=(5)/(16)`
`therefore (3x+1)/((x-2)^(2)(x+2))=(((5)/(16)))/(x-2)+(((7)/(4)))/((x-2)^(2))+((-(5)/(16)))/(x+2)`
`=(5)/(16)int (1)/(x-2)dx+(7)/(4)int(x-2)^(-2)dx-(5)/(16)int(1)/(x+2)dx`
`therefore =int [(((5)/(16)))/(x-2)+(((7)/(4)))/((x-2)^(2))+((-(5)/(16)))/(x+2)]dx`
`=(5)/(16)int (1)/(x-2)dx+(7)/(4)int(x-2)^(-2)dx-(5)/(16)int(1)/(x+2)dx`
`=(5)/(16)log|x-2|+(7)/(4).((x-2)^(-1))/(-1).(1)/(1)-(5)/(16)log|x+2|+c`
`=(5)/(16)log|(x-2)/(x+2)|-(7)/(4(x-2))+c`