Here,
I=int(2x+3)/(x^4-9x^2)dx=int(2x+3)/(x^2(x^2-9))dx
I=int(2x+3)/(x^2(x-3)(x+3))dx
Let,
(2x+3)/(x^2(x-3)(x+3))=A/x+B/x^2+C/(x-3)+D/(x+3)
2x+3=Ax(x^2-9)+B(x^2-9)+Cx^2(x+3)+Dx^2(x-3)
2x+3=x^3(A+C+D)+x^2(B+3C-3D)+x(-9A)-9B
"Comparing co efficients of " x^3,x^2,x and "constant term
:"
A+C+D=0..........to(1)
B+3C-3D=0.......to(2)
-9A=2=>color(red)(A=-2/9to(3)
-9B=3=>color(red)(B=-1/3to(4)
From (1) and (3)color(white)(..........)From (2) and(4)
C+D=2/9to(5)color(white)(.......)3C-3D=1/3=>C-D=1/9to(6)
Adding (5) and (6) we get
C+C=2/9+1/9=>2C=3/9=>C=3/18=>color(red)(C=1/6
From (5) ,we get
1/6+D=2/9=>D=2/9-1/6=(4-3)/18=>color(red)(D=1/18
So,
I=int[(-2/9)/x+(-1/3)/x^2+(1/6)/(x-3)+(1/18)/(x+3)]dx
=-2/9ln|x|-1/3(x^(-2+1)/(-2+1))+1/6ln|x-3|+1/18ln|x+3| + c
=-2/9ln|x|-1/3(x^(-1)/(-1))+1/6ln|x-3|+1/18ln|x+3|+c
=1/18ln|x+3|+1/6ln|x-3|-2/9ln|x|+1/(3x)+c
