what is the integral of sqrt(1+4x^2)dx between 0 and 2 using teh substitution x = (1/2)sinh(theta)(hyperbolic)

Answer 1

Is it possible that required integral is:-
I = ∫ 1 / (1 + 4x²)^(1/2) dx
Let u = 2x
du = 2 dx
I = (1/2) ∫ [1 / (1 + u²) ^(1/2) ] du
I = (1/2) sinh ^(-1) u
I = (1/2) sinh^(-1) (2x)
I = (1/2) [sinh^(-1) 4 - sinh^(-1) 0 ]
I = (1/2) sinh^(-1) 4