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

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How do you show whether the improper integral $\int (79 \frac{x^{2}}{9 + x^{6}}) d x$ $int (79 x^2/(9 + x^6)) dx$ 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.

Explanation:

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

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

As $x \to \infty$ $xrarroo$ , we have ${tan}^{- 1} (\frac{x^{3}}{3}) \to \frac{π}{2}$ $tan^-1(x^3/3) rarr pi/2$ (and as $x \to - \infty$ $xrarr-oo$ , we have ${tan}^{- 1} (\frac{x^{3}}{3}) \to - \frac{π}{2}$ $tan^-1(x^3/3) rarr -pi/2$ ) so both

$\int_{- \infty}^{0} (\frac{x^{2}}{9 + x^{6}}) d x$ $int_-oo^0 (x^2/(9 + x^6)) dx$ and $\int_{0}^{\infty} (\frac{x^{2}}{9 + x^{6}}) d x$ $int_0^oo (x^2/(9 + x^6)) dx$ converge.

Therefore, $\int_{- \infty}^{\infty} (\frac{x^{2}}{9 + x^{6}}) d x$ $int_-oo^oo (x^2/(9 + x^6)) dx$ converges.

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

1 Answer

Explanation:

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