integral of 2/(4x^2+1)=?

Question

) 2/(4x^2+1)=?
(

Answer 1

2 / (4x² + 1)
= 2 / 4(x² + 1/4)
I = (1/2) ∫ 1 / (x² + (1/2)² ) dx
I = (1/2)(2) tan ^(-1) (2x) + C
I = tan^(-1) (2x) + C

Answer 2

Int 2dx /(4x^2+1)
let 2x = tan u
then
you are looking at a right angled traingle
with sides 2x and 1 and hypothenuse sqrt (4x^2 +1)
2x = tan u
2dx = sec^2 u du
4x^2 +1 = cos^2 u

Int 2dx /(4x^2+1) = Int sec^2 u du cos^2 u
= Int du = u
= tan^-1 (2x)

Answer 3

∫2/(4x^2+1) dx
= ∫1/[(2x)^2+1)] d(2x)
= arctan(2x) + c