Find ∫ 1/(1-4x^2) dx with a low limit of 0 and a high limit of 1

Find ∫ 1/(1-4x^2) dx with a low limit of 0 and a high limit of 1 (another way to put it is Integrate[1/(1 - 4 x^2), {x, 0, 1}]

Anyway this is what I did:

I knew that this definite integral was = 1/2 tan^-1 (2) which equal about 1.553574...

I got the above from knowing that the integral of 1/(4x^2 )+1 is 1/2 tan ^-1 (2x)

=1/2 tan ^-1 (2x) + constant

(what I want to know is if this is enough work/steps to show how I got the answer- if it is not could someone tell what steps I would have to show- Thanks for the help)