Question 12352
 {{{(x+2)(x+2)/(x+1)<0}}}
 is {{{ (x+2)^2/(x+1) < 0 }}}
 Hence,the numberator and the denominator are of different sign.
 But {{{ (x+2)^2 >= 0 }}} for all real x.
 
 The denomiator {{{ x+1 }}} must be negative and the nmerator
 {{{x+2 }}} cannot be 0.
 
 So, {{{ (x+2)^2/(x+1) < 0 }}} implies
 {{{ x+1<0 and x is not equal to -2}}}
  
 We obtain the solution set {{{ -2 < x < -1 or x < -2}}}

 In the interval notaion: (-oo,-2) U (-2,-1).

 The above is the shortest way to solve this kind of questions.
 Try to think about it. [In fact, I can get the answer by direct view
 in a second. ]

 If you really want to consider cases

 Look the sign of the given  {{{(x+2)^2/(x+1)}}} on the intervals
 (-oo,-2),(-2,-1), (-1,+oo). You will see.
 
           
 -----------*-----*----------------------
 -oo        -2    -1                  +oo


 Kenny