Integral Convergence/Divergence: 0 to ∞, 1/(1+x^6)^(1/2)

Lchan1
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Homework Statement


Determine if the integral converges or diverges?
it;s the integral of 0 to infinity
of 1/(1+x^6)^(1/2)

Homework Equations



so I compared it with 1/x^2

The Attempt at a Solution



the answer key says it converges but i think it diverges since the integral of 1/x^2 diverges from 0 to 1...
 
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1/x^2 is WAY GREATER than 1/(1+x^6)^(1/2) near 0. In fact, the latter function is bounded on [0,1]. The fact 1/x^2 diverges near zero doesn't prove your function does.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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