Question

Question #76899

Answer 1

It converges. Compare it with `1/x^3` for `x>=1`.

Explanation:

I'm going to compare `1/sqrt(x^6+1)` to `1/x^3`. However, note that `1/x^3` is not defined at `x=0`. Therefore, I'm going to start the integral at `x=1` instead.

`int_1^oo 1/sqrt(x^6+1) dx`

Since `0 < 1/sqrt(x^6+1) < 1/x^3` for all `x>=1`, and

`int_1^oo 1/x^3 dx = 1/2`

The integral

`int_1^oo 1/sqrt(x^6+1) dx`

must therefore converge too.

The integral

`int_0^1 1/sqrt(x^6+1) dx`

converges as it is a proper integral. Putting the 2 integrals together

`int_0^1 1/sqrt(x^6+1) dx + int_1^oo 1/sqrt(x^6+1) dx = int_0^oo 1/sqrt(x^6+1) dx`

The resultant integral converges as well.