If (2 x+4) x-1 ~d x= a (x-1)^ 5 2 + b (x-1)^ 3 2 + c , (where c is a constant of integration), then the value of a+b is

Step-by-step solution

array ll & Let I= (2 x+4) x-1 d x & Put x-1=t^2 & d x=2 t d t & I= (2 (t^2+1 )+4 ) t 2 t d t array aligned & = (2 t ^2+2+4 ) 2 t ^2 dt & = (2 t ^2+6 ) 2 t ^2 dt & = (4 t ^4+12 t ^2 ) dt & =4 t ^4 dt +12 t ^2 dt & =4 t ^5 5 +12 t ^3 3 + C & = 4 5 t ^5+4 t ^3+ C & = 4 5 (x-1)^ 5 2 +4(x-1)^ 3 2 +C aligned But (2 x+4) x-1 ~d x= a (x-1)^ 5 2 + b (x-1)^ 3 2 + C aligned & a= 4 5 and b=4 & a+b= 4 5 +4= 24 5 aligned