How do you integrate : taninverse(2x) / (1+4x²)

∫ arctan(2x)/(1 + 4x²) dx

Let u = arctan(2x). Then,

du/dx = 2/(1 + (2x)²) [Note that d/dx arctan(f(x)) = f'(x)/(1 + f(x)²)]
du/dx = 2/(1 + 4x²)
du = 2/(1 + 4x²) dx
du/2 = dx/(1 + 4x²) [Rearrange the terms to make dx/(1 + 4x²) the subject]

So:

∫ u (dx/(1 + 4x²))
= ∫ u du/2
= ½ ∫ u du

Finally, by power rule ∫ xⁿ dx = x^(n + 1)/(n + 1) + c:

½u^(1 + 1)/(1 + 1) + c
= ½ * u²/2 + c
= u²/4 + c
= (arctan(2x))²/4 + c [Substitute u back with arctan(2x)]

I hope this helps!

well you can't use the tan inverse on this problem. this could solve this with logarithmic formula.
....arctan(2x)dx
∫--------------
....1+4x^2



let u=arctan(2x)
...........1
du = -------------- d/dx 2 dx
........1+(2x)^2

...........2dx
du =---------------
........1+4x^2


=1/2 ∫ udu
.....u^2
=-----------+C
.....4
substitute back for u= tan^(-1)(2x)


= 1/4tan^(-1)^2(2x) +C answer//

Choose u = tan(2x) then du = 2dx/(1+4x^2). Then you can rewrite the integrand as
int(u/2, u) = 1/4 * u^2 = 1/4 [arctan(2x)]^2 + C