`(x+2)/(x(x^2-9))`
`(x+2)/(x(x^2-9))=(x+2)/(x(x+3)(x-3))`
Now let `(x+2)/(x(x^2-9))=A/x+B/(x+3)+C/(x-3)`
`(x+2)/(x(x^2-9))=(A(x+3)(x-3)+B(x)(x-3)+C(x)(x+3))/(x(x+3)(x-3))`
`(x+2)/(x(x^2-9))=(A(x^2-9)+B(x^2-3x)+C(x^2+3x))/(x(x^2-9))`
`:.(x+2)=A(x^2-9)+B(x^2-3x)+C(x^2+3x)`
`x+2=Ax^2-9A+Bx^2-3Bx+Cx^2+3Cx`
`x+2=(A+B+C)x^2+(-3B+3C)x-9A`
equating the coefficients of the like terms,
`A+B+C=0`
`-3B+3C=1`
`-9A=2`
Solve the above three equations to get the values of A,B and C,
`A=-2/9`
Bach substitute the value of A in the first equation,
`-2/9+B+C=0`
`B+C=2/9`
From the above equation ,express C in terms of B
`C=2/9-B`
Substitute the expression of C in the second equation,
`-3B+3(2/9-B)=1`
`-3B+2/3-3B=1`
`-6B=1-2/3`
`-6B=1/3`
`B=-1/18`
Now plug the value of...
`(x+2)/(x(x^2-9))`
`(x+2)/(x(x^2-9))=(x+2)/(x(x+3)(x-3))`
Now let `(x+2)/(x(x^2-9))=A/x+B/(x+3)+C/(x-3)`
`(x+2)/(x(x^2-9))=(A(x+3)(x-3)+B(x)(x-3)+C(x)(x+3))/(x(x+3)(x-3))`
`(x+2)/(x(x^2-9))=(A(x^2-9)+B(x^2-3x)+C(x^2+3x))/(x(x^2-9))`
`:.(x+2)=A(x^2-9)+B(x^2-3x)+C(x^2+3x)`
`x+2=Ax^2-9A+Bx^2-3Bx+Cx^2+3Cx`
`x+2=(A+B+C)x^2+(-3B+3C)x-9A`
equating the coefficients of the like terms,
`A+B+C=0`
`-3B+3C=1`
`-9A=2`
Solve the above three equations to get the values of A,B and C,
`A=-2/9`
Bach substitute the value of A in the first equation,
`-2/9+B+C=0`
`B+C=2/9`
From the above equation ,express C in terms of B
`C=2/9-B`
Substitute the expression of C in the second equation,
`-3B+3(2/9-B)=1`
`-3B+2/3-3B=1`
`-6B=1-2/3`
`-6B=1/3`
`B=-1/18`
Now plug the value of B in the expression of C,
`C=2/9-(-1/18)`
`C=2/9+1/18`
`C=(2*2+1)/18`
`C=5/18`
`:.(x+2)/(x(x^2-9))=-2/(9x)-1/(18(x+3))+5/(18(x-3))`
Now let's check the result,
RHS=`-2/(9x)-1/(18(x+3))+5/(18(x-3))`
`=(-2*2(x+3)(x-3)-x(x-3)+5x(x+3))/(18x(x+3)(x-3))`
`=(-4(x^2-9)-(x^2-3x)+5(x^2+3x))/(18x(x^2-9))`
`=(-4x^2+36-x^2+3x+5x^2+15x)/(18x(x^2-9))`
`=(18x+36)/(18x(x^2-9))`
`=(18(x+2))/(18x(x^2-9))`
`=(x+2)/(x(x^2-9))`
=LHS
Hence it is verified.
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