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

Mar 18, 2016

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

#### Explanation:

We can take out a constant factor, so ${\int}_{-} {\infty}^{\infty} \left(79 {x}^{2} / \left(9 + {x}^{6}\right)\right) \mathrm{dx}$ converges if and only if ${\int}_{-} {\infty}^{\infty} \left({x}^{2} / \left(9 + {x}^{6}\right)\right) \mathrm{dx}$ converges.

$\int \left({x}^{2} / \left(9 + {x}^{6}\right)\right) \mathrm{dx} = \frac{1}{9} {\tan}^{-} 1 \left({x}^{3} / 3\right)$

As $x \rightarrow \infty$, we have ${\tan}^{-} 1 \left({x}^{3} / 3\right) \rightarrow \frac{\pi}{2}$ (and as $x \rightarrow - \infty$, we have ${\tan}^{-} 1 \left({x}^{3} / 3\right) \rightarrow - \frac{\pi}{2}$) so both

${\int}_{-} {\infty}^{0} \left({x}^{2} / \left(9 + {x}^{6}\right)\right) \mathrm{dx}$ and ${\int}_{0}^{\infty} \left({x}^{2} / \left(9 + {x}^{6}\right)\right) \mathrm{dx}$ converge.

Therefore, ${\int}_{-} {\infty}^{\infty} \left({x}^{2} / \left(9 + {x}^{6}\right)\right) \mathrm{dx}$ converges.