`f(x)=((x+1)(x+2))/((x-1)(x-2))`
`f(x)=0. thereforex=-1, -2.`
Also, f(x) is discontinuous at x=1 and x=2.
`underset(xrarrI^(-))limf(x)rarr+ooand underset(xrarrI^(+))limf(x)rarr-oo`
`underset(xrarr2^(-))limf(x)rarr-ooand underset(xrarr2^(+))limf(x)rarr+oo,f(0)=1`
`underset(xrarrpmoo)limf(x)rarr1`
So, graph of the function is as shown in the following figure.
The area of the shaded region is
`|underset(-2)overset(-1)intf(x)dx|=|underset(-2)overset(-1)int((x^(2)+3x+2)/(x^(2)-3x+2))dx|`
`=|underset(-2)overset(-1)int1+(6x)/((x-1)(x-2))dx|`
`=|underset(-2)overset(-1)int[1+6((2)/(x-2)-(1)/(x-1))]dx|`
`=|[x+12 log_(e)|x-2|-6 log_(e)|x-1|]_(-2)^(-1)|`
`=|[1+12 log_(e)""(3)/(4)-6 log_(e)""(2)/(3)]|`
`=|[1+6 log_(e)""(9)/(16)-6log_(e)""(2)/(3)]|`
`=|[1+6 log_(e)""(27)/(32)]|`
`=6 log_(e)""(32)/(27)-1` sq. units
