the SM shows a different method but i figure i should arrive at the same answer either way. what step am i screwing up on? thanks!(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_{ - \infty}^{\infty}\frac {x^{2}dx}{9 + x^6}[/tex]

[tex]\int_{ - \infty}^{0}\frac {x^{2}dx}{9 + x^6} + \int_{0}^{\infty}\frac {x^{2}dx}{9 + x^6}[/tex]

[tex]u = x^{3}[/tex]

[tex]du = 3x^{2}dx[/tex]

[tex]\frac {1}{3}[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{9 + u^{2}} + \lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{9 + u^{2}}][/tex]

[tex]3[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} + \lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{1 + (\frac {u}{3})^{2}}][/tex]

[tex]3[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{0} - \arctan{\frac {t^{3}}{3}] +

3[\lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{\frac {t^{3}}{3} - \arctan{0}][/tex]

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# Homework Help: Where am i screwing up? Improper Integral

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