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?
Is there a quicker way of doing this integral?
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1Or better: [this one](http://math.stackexchange.com/a/43466) – 2012-05-17
2 Answers
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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0Not at all, Ray – 2012-05-17
$\int_0^\infty\frac{x^2}{1+x^4}dx$
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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0aha 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