`(x^2 + 5)/((x + 1)(x^2 - 2x + 3))` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

`(x^2+5)/((x+1)(x^2-2x+3))` 

Let`(x^2+5)/((x+1)(x^2-2x+3))=A/(x+1)+(Bx+C)/(x^2-2x+3)`

`(x^2+5)/((x+1)(x^2-2x+3))=(A(x^2-2x+3)+(Bx+C)(x+1))/((x+1)(x^2-2x+3))`

`(x^2+5)/((x+1)(x^2-2x+3))=(Ax^2-2Ax+3A+Bx^2+Bx+Cx+C)/((x+1)(x^2-2x+3))`

`:.(x^2+5)=Ax^2-2Ax+3A+Bx^2+Bx+Cx+C`

`x^2+5=(A+B)x^2+(-2A+B+C)x+3A+C`

equating the coefficients of the like terms,

`A+B=1`

`-2A+B+C=0`

`3A+C=5`

Now let's solve the above three equations to find the values of A,B and C,

Express C in terms of A from the third equation,

`C=5-3A`

Substitute the above expression of C in second equation,

`-2A+B+5-3A=0`

`-5A+B+5=0`

`-5A+B=-5`

Now subtract the first equation from the above equation,

`(-5A+B)-(A+B)=-5-1`

`-6A=-6`

`A=1`

Plug the value of A in the first and...

`(x^2+5)/((x+1)(x^2-2x+3))` 

Let`(x^2+5)/((x+1)(x^2-2x+3))=A/(x+1)+(Bx+C)/(x^2-2x+3)`

`(x^2+5)/((x+1)(x^2-2x+3))=(A(x^2-2x+3)+(Bx+C)(x+1))/((x+1)(x^2-2x+3))`

`(x^2+5)/((x+1)(x^2-2x+3))=(Ax^2-2Ax+3A+Bx^2+Bx+Cx+C)/((x+1)(x^2-2x+3))`

`:.(x^2+5)=Ax^2-2Ax+3A+Bx^2+Bx+Cx+C`

`x^2+5=(A+B)x^2+(-2A+B+C)x+3A+C`

equating the coefficients of the like terms,

`A+B=1`

`-2A+B+C=0`

`3A+C=5`

Now let's solve the above three equations to find the values of A,B and C,

Express C in terms of A from the third equation,

`C=5-3A`

Substitute the above expression of C in second equation,

`-2A+B+5-3A=0`

`-5A+B+5=0`

`-5A+B=-5`

Now subtract the first equation from the above equation,

`(-5A+B)-(A+B)=-5-1`

`-6A=-6`

`A=1`

Plug the value of A in the first and third equation to get the values of B and C,

`1+B=1`  

`B=1-1`

`B=0`

`3(1)+C=5`

`C=5-3`

`C=2`

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

Now let's check it algebraically,

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

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

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

Hence it is verified.