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I have an integral $\int_0^\infty {x^2\over 1+x^4} dx$. I gave it a go and it turned out quite messy, so I consulted Wolfram Alpha but the steps given there seem rather long winded too. Is there is a faster way of doing the integral?

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    Or better: [this one](http://math.stackexchange.com/a/43466)2012-05-17

2 Answers 2

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As SauravTomar pointed out $ \int\limits_{0}^\infty\frac{x^2}{x^4+1}dx=\int\limits_{0}^\infty\frac{1}{x^4+1}dx $ so $ \int\limits_{0}^\infty\frac{x^2}{x^4+1}dx= \frac{1}{2}\int\limits_{0}^\infty\frac{x^2+1}{x^4+1}dx= \frac{1}{2}\int\limits_{0}^\infty\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx $ $ =\frac{1}{2}\int\limits_{0}^\infty\frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+2}dx= \frac{1}{2}\int\limits_{-\infty}^\infty\frac{dt}{t^2+2}dx= \frac{\pi}{2\sqrt{2}}. $

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    Not at all, Ray2012-05-17
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$\int_0^\infty\frac{x^2}{1+x^4}dx$

Let, $t=\frac{1}{x}$, and, $dt=\frac{-dx}{x^2}$

$\int_\infty^0 \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$

$=\int_\infty^0\frac{-dt}{1+t^4} =\int_0^\infty \frac{dt}{1+t^4}$

Follow Norberts solution after that.


Another way to do this is

$ I=\int_0^\infty\frac{x^2}{1+x^4}dx $ Let, $x= \sqrt{\tan\theta}$, then $dx=\frac{1}{2\sqrt{\tan\theta}}\sec^2\theta d\theta$

$ I=\int_0^{\frac{\pi}{2}} \frac{\tan\theta}{1+\tan^2\theta}\times\frac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$

$ I=\frac{1}{2}\times\int_0^{\frac{\pi}{2}}\sqrt{\tan\theta} $ also, $ I=\frac{1}{2}\times\int_0^{\frac{\pi}{2}}\sqrt{\cot\theta} $ hence, $ 4I=\int_0^{\frac{\pi}{2}}\sqrt{\tan\theta}+\sqrt{\cot\theta} $

$ 4I=\int_0^{\frac{\pi}{2}} \frac{\sin\theta + \cos\theta}{\sqrt{\sin\theta\cos\theta}} $

$=\sqrt2 \int_0^{\frac{\pi}{2}} \frac{(\sin\theta + \cos\theta)}{\sqrt{1-(\sin\theta - \cos\theta)^2}}$

Let $t=\sin\theta - \cos\theta$, then

$4I=\sqrt2 \int_{-1}^{1} \frac{dt}{\sqrt{1-t^2}}$

$I=\frac{1}{2\sqrt2}\left(\sin^{-1}(1)-\sin^{-1}(-1)\right) =\frac{\pi}{2\sqrt2}$

:) $ \int_0^1 \left(\sqrt[3]{1-x^7} \right)$

$\frac{a}{d}$

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    aha got it, then i guess i will try it di$f$$f$erent way,(similar to norberts though, but a little bit di$f$ferent)2012-05-17