Can someone help me plz with this integral?

Question

Evaluate the integral from 6 to infinity of
dx/(x^(2/3) + 1)

Answer 1

Knowing that ∫(du / (u² + 1) = arctan(u), we do some convoluted manipulation:
dx / (x^(⅔) + 1) = (mulitplying by x^(-⅔) * x^(⅔)
(x^(-⅔) dx) * (x^(⅔)/(x^(⅔) + 1)

Let u = x^(⅓).
Then du = ⅓x^(-⅔)dx, or 3du = x^(-⅔)dx

Substituting for u and du, we have:
∫3du * (u² / (u² + 1)) =
∫3((u² + 1 - 1) / (u² + 1))du =
∫3((u² + 1)/(u² + 1) - 1/(u² + 1))du =
∫3(1 - 1/(u² + 1))du

breaking this up and integrating:
∫3du - ∫3du/(u² + 1) =
3∫du - 3∫du/(u² + 1) =
3u - 3arctan(u)

Upper bound:
x→∞
u = x^(⅓)
u→∞

Therefore, at the upper bound, 3u - 3arctan(u) also tends to infinity, indicating that the function does not converge fast enough to have a finite integral over this interval.

Doesn't seem reasonable? Consider that 1/(x^(⅔) + 1) grows at almost eactly the same rate as 1/(x^(⅔)),
the indefinite integral of which is 3x^(⅓) + C, which also grows without bounds despite the function itself tending toward 0.

Answer 2

^ (1/3) ^ (1/3) ^1/2
arctan(x) - 2 arctan(x ) - arctan(2 x + 3 )

^(1/3) ^1/2 ^ (1/3)
+ arctan(-2 x + 3 ) + 3 x