₁^2 d x (x^2-2 x+4 )^ 3 2 = k k +5 , then k has the value

Step-by-step solution

Let aligned I & = ₁^2 d x (x^2-2 x+4 )^ 3 2 & = ₁^2 d x [(x-1)^2+3 ]^ 3 2 aligned Put x-1= 3 d x= 3 ^2 d When x=1, =0 When x=2, = 6 aligned & I= ₀^ 6 3 ^2 [3 ^2 +3 ]^ 3 2 d & = ₀^ 6 3 ^2 [3 (1+ ^2 ) ]^ 3 2 & = ₀^ 6 3 ^2 3 3 ( ^2 )^ 3 2 & = ₀^ 6 1 3 ^2 ^3 & = 1 3 _ ^ 6 & = 1 3 [ ]₀^ 6 & I= 1 3 [ 6 - 0 ] & I= 1 6 & k k+5 = 1 6 & 6 k=k+5 & k =1 & aligned