Function y = 2 ^ (x + 2) - 3x4 ^ x, if x ^ 2 + X

Function y = 2 ^ (x + 2) - 3x4 ^ x, if x ^ 2 + X

a=2^x
x(x+1)

∫ 1 / X (x ^ 2 + 1) ^ (1 / 2) DX =? Please talk about the process,

∫ dx/[x√(x^2+1)]
let
x= tany
dx = (secy)^2dy
∫ (secy/tany) dy
=∫ cscy dy
=ln|cscy-coty| + C
=ln|√(x^2+1)/x - 1/x | + C

∫dx/x*(x^2-1)^1/2
1、 Let x = sectdx = SiNx / (cosx) ^ 2 DT (x ^ 2-1) = (sect) ^ 2-1 = (TaNx) ^ 2 radical (x ^ 2-1) = tanxdx / (x * radical (x ^ 2-1)) = DX / (sect * tant) = DX / (1 / cost) * TaNx = acecos (1 / x)
2、 Let t = 1 / x, then DX = - DT / T ^ 2 ∫ 1 / [x (x ^ 2-1) ^ (1 / 2)] DX = - ∫ (DT / T ^ 2) * t | t | / (1-T ^ 2) = - SGN (T) ∫ DT / (1-T ^ 2) ^ (1 / 2) = - SGN (x) arcsint + C = - arcsin (1 / | x | + C
I think both methods are right. Why are the answers different? Just teach

The two answers are equivalent because the sum of arcsin and arcsin of the same angle is constant: arcsin a + arccos a = π / 2, so arccos (1 / x) + arcsin (1 / x) = π / 2arccos (1 / x) = - arcsin (1 / x) + π / 2. They are equivalent because they differ by a constant