If, d ^2 ( 2 + 2 ) = +2 _ e | f ( )|+ c (where c is a constant of integration), then the ordered pair ( ,|f( )|) is equal to

Step-by-step solution

Let aligned I & = d ^2 ( 2 + 2 ) & = ^2 d ( 2 1- ^2 )+ ( 1+ ^2 1- ^2 ) & = ^2 (1- ^2 ) d (1+ )^2 aligned aligned & Put =t ^2 d =d t I & = (1-t^2 ) d t (1+t)^2 = 1-t 1+t d t & = 2-(1+t) 1+t d t & =2 |1+t|-t+C aligned =2 |1+ |- +C Comparing with +2 |f( )|+C , we get =-1, f ( )=1+