Question

How do you find the indefinite integral of `int (x^2-3)/((x^2-9)(x+2)^2) dx`?

Answer 1

`int(x^2-3)/((x^2-9)*(x+2)^2)dx=-log(x+3)+log(x-3)/25+(19/25)*log(x+2)+1/(5(x+2))+C`

Explanation:

let us assume `(x^2 -3)/((x^2-9)*(x+2)^2)=a/(x+3)+b/(x-3)+c/(x+2)+d/(x+2)^2`

then

`(x^2 -3)=a*(x-3)*(x+2)^2+b*(x+3)*(x+2)^2+c*(x^2-9)*(x+2)+d*(x^2-9)`

put x=3 in above equation
then
`6=b*6*25+a*0+c*0+d*0`
`b=1/25`
now
put x=-3 then
`6=a*(-6)+b*0+c*0+d*0`
`a=-1`

put x=-2
`1=a*0+b*0+c*5*0+d*5`

`d=1/5`

put x=0

`(-3)=a*(-3)*4+b*3*4+c*(-9)*2+d*(-9)`

now let us put the value of a, b and d in above equation to find the value of c

`(-3)=(-12)*(-1)+(1/25)*12+(-18)*c+(1/5)*(-9)`

`c=19/25`

therefore

`int(x^2 -3)/((x^2-9)*(x+2)^2)dx = int((-1)/(x+3)+1/(25*(x-3))+19/(25*(x+2))+1/(5*(x+2)^2)) dx`

`int(x^2-3)/((x^2-9)*(x+2)^2)dx=-log(x+3)+log(x-3)/25+(19/25)*log(x+2)+1/(5(x+2))+C`