Question #76899

Nov 18, 2017

It converges. Compare it with $\frac{1}{x} ^ 3$ for $x \ge 1$.

Explanation:

I'm going to compare $\frac{1}{\sqrt{{x}^{6} + 1}}$ to $\frac{1}{x} ^ 3$. However, note that $\frac{1}{x} ^ 3$ is not defined at $x = 0$. Therefore, I'm going to start the integral at $x = 1$ instead.

${\int}_{1}^{\infty} \frac{1}{\sqrt{{x}^{6} + 1}} \mathrm{dx}$

Since $0 < \frac{1}{\sqrt{{x}^{6} + 1}} < \frac{1}{x} ^ 3$ for all $x \ge 1$, and

${\int}_{1}^{\infty} \frac{1}{x} ^ 3 \mathrm{dx} = \frac{1}{2}$

The integral

${\int}_{1}^{\infty} \frac{1}{\sqrt{{x}^{6} + 1}} \mathrm{dx}$

must therefore converge too.

The integral

${\int}_{0}^{1} \frac{1}{\sqrt{{x}^{6} + 1}} \mathrm{dx}$

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

${\int}_{0}^{1} \frac{1}{\sqrt{{x}^{6} + 1}} \mathrm{dx} + {\int}_{1}^{\infty} \frac{1}{\sqrt{{x}^{6} + 1}} \mathrm{dx} = {\int}_{0}^{\infty} \frac{1}{\sqrt{{x}^{6} + 1}} \mathrm{dx}$

The resultant integral converges as well.