How to integrate (2t^2) / (2t^2 + 1)?

Answer 1

∫{2t²/(2t² + 1)}dt
∫{t²/(t² + ½)}dt
Make a trigonometric substitution
t = [√(2)/2] tan Θ
dt = [√(2)/2] (secΘ)² dΘ
Substitute into the integral
½∫{tan² Θ/½(tan²Θ+ 1)} [√(2)/2] (secΘ)² dΘ
∫{tan² Θ} [√(2)/4] dΘ
[√(2)/4]{tan Θ - Θ} + C
but remember that t = [√(2)/2] tan Θ
[√(2)/4][t/[√(2)/2] - arctan{t/[√(2)/2]} + C
2t/8 - [√(2)/4][arctan{t/[√(2)/2]} + C

Answer 2

(2t^2) / (2t^2 + 1) = 1 + 1/2t^2 = 1 + t^-2/2
Now you should use power rule to integrate each term and don't for get to add a constant at the end.

Answer 3

t+1 that is if by / you mean divide
lol took me like 1 minute to do that, i hate that kinda problem, its harder than the rest