If x+1 2 x-1 ~d x= f (x) 2 x-1 + c , (where c is a constant of integration), then f (x) is equal to

Step-by-step solution

Let I = x+1 2 x-1 ~d x Put 2 x-1= t ^2 x +1= t ^2+3 2 aligned & d x=t dt & = ( t ^2+3 2 ) tdt t & = t ^3 6 + 3 2 t + c & = t 2 ( t ^2+9 3 )+ c & = 2 x-1 2 ( 2 x-1+9 3 )+ c & = 2 x-1 ( x+4 3 )+ c aligned Comparing with f (x) 2 x-1 + c , we get f (x)= x+4 3