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

Question

hi .C-wryte olease solve this i am sure you will solve it and my id ali_kh144 from iran

Answer 1

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

Answer 2

Int 1/(4x^2 +1) dx
= Int 1/((2x)^2 + 1) dx
u = 2x
du = 2dx
= (1/2) Int 1/(u^2 + 1) du
= (1/2) Arctan(u) + C
= (1/2) Arctan(2x) + C

Answer 3

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

We know that ∫ 1/(u^2 + 1) du = arctan u + C

Thus, let u = 2x
du/dx = 2
dx = du/2

This gives,

∫ 1/(u^2+1) du/2

(1/2) arctan(u) + C

But u = 2x

(1/2) arctan(2x) + C

Answer 4

u = 2x
du = 2dx
du/2 = dx
∫1/(u²+1) du/2
1/2 ∫1/(u²+1) du
1/2 * arctan(u) + C

1/2 * arctan(2x) + C

Answer 5

Use trig substitution.