How do you show whether the improper integral $int (79 x^2/(9 + x^6)) dx$ converges or diverges from negative infinity to infinity?

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Answer 1 · 2016-03-18T13:25:24

I would integrate by trigonometric substitution, then check that the limit exists.

We can take out a constant factor, so $int_-oo^oo (79x^2/(9 + x^6)) dx$ converges if and only if $int_-oo^oo (x^2/(9 + x^6)) dx$ converges.

$int (x^2/(9 + x^6)) dx = 1/9tan^-1(x^3/3)$

As $xrarroo$ , we have $tan^-1(x^3/3) rarr pi/2$ (and as $xrarr-oo$ , we have $tan^-1(x^3/3) rarr -pi/2$ ) so both

$int_-oo^0 (x^2/(9 + x^6)) dx$ and $int_0^oo (x^2/(9 + x^6)) dx$ converge.

Therefore, $int_-oo^oo (x^2/(9 + x^6)) dx$ converges.

How do you show whether the improper integral int (79 x^2/(9 + x^6)) dxint (79 x^2/(9 + x^6)) dx converges or diverges from negative infinity to infinity?