Question #76899

1 Answer
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Nov 18, 2017

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.

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