Finding Convergence or Divergence of $$\int^{\infty}_{0}\frac{1}{\sqrt{x^6+1}}dx$$
What i Try:
I am trying to prove $(x^2+1)\leq \sqrt{x^6+1}$ for all $x\geq 0$
$(x^2+1)^2\leq (x^6+1)\Longrightarrow x^4+2x^2+1\leq x^6+1$
Getting $x^4+2x^2\leq x^6$ but which is false.
How do i finding that integral is converges. Help me Thanks
$$(\forall x\ge 1)\;\; x^6+1>x^6$$
$$\implies \; (\forall x\ge 1)\;\; \sqrt{x^6+1}> \sqrt{x^6}$$
$$\implies \;\;(\forall x\ge 1)\;\; \frac{1}{\sqrt{x^6+1}}<\frac{1}{x^3}$$
but $\int_1^{+\infty}\frac{dx}{x^3} $ converges, thus, by comparison criterion, $\int_1^{+\infty}\frac{dx}{\sqrt{x^6+1}}$ converges and finally
$$\int_0^{+\infty}\frac{dx}{\sqrt{x^6+1}} \;\; is \;\; convergent$$
