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

1 Answer
Sep 23, 2015

#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#