`(x+1)/(x^2-x-6)`

Let's factorize the denominator,

`x^2-x-6=x^2-3x+2x-6`

`=x(x-3)+2(x-3)`

`=(x-3)(x+2)`

`:.(x+1)/(x^2-x-6)=(x+1)/((x-3)(x+2))`

Now let`(x+1)/(x^2-x-6)=A/(x-3)+B/(x+2)`

`(x+1)/(x^2-x-6)=(A(x+2)+B(x-3))/((x-3)(x+2))`

`(x+1)/(x^2-x-6)=(Ax+2A+Bx-3B)/((x-3)(x+2))`

`:.(x+1)=Ax+2A+Bx-3B`

`x+1=(A+B)x+2A-3B`

Equating the coefficients of the like terms,

`A+B=1`

`2A-3B=1`

Now let's solve the above two equations to get the values of A and B,

express B in terms of A from the first equation,

`B=1-A`

substitute the expression of B in the second equation,

`2A-3(1-A)=1`

`2A-3+3A=1`

`5A-3=1`

`5A=1+3`

`5A=4`

`A=4/5`

Plug the value of A in the first equation,

`4/5+B=1`

`B=1-4/5`

`B=1/5`

`:.(x+1)/(x^2-x-6)=4/(5(x-3))+1/(5(x+2))`

Now let's check the above result,

RHS=`4/(5(x-3))+1/(5(x+2))`

`=(4(x+2)+1(x-3))/(5(x-3)(x+2))`

`=(4x+8+x-3)/(5(x^2+2x-3x-6))`

`=(5x+5)/(5(x^2-x-6))`

`=(5(x+1))/(5(x^2-x-6))`

`=(x+1)/(x^2-x-6)`

= LHS

Hence it is verified.

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