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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    $\begingroup$ I am incompetent at searching, but I remember this problem from MSE, and seem to recall giving a solution. $\endgroup$ – André Nicolas May 17 '12 at 14:36
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    $\begingroup$ @André: I think you might have this thread in mind. $\endgroup$ – t.b. May 17 '12 at 14:44
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    $\begingroup$ Or better: this one $\endgroup$ – t.b. May 17 '12 at 14:58

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

  • $\begingroup$ May I ask what were the tell-tale signs of the original integrand that helped you spot this substitution? $\endgroup$ – Ray May 17 '12 at 15:34
  • $\begingroup$ I saw this trick several times $\endgroup$ – Norbert May 17 '12 at 15:41
  • $\begingroup$ Not at all, Ray $\endgroup$ – Norbert May 17 '12 at 15:54


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

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

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

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


  • $\begingroup$ -1 Please Norbert's solution. You initial steps are valid, but their following is not. $\endgroup$ – Sasha May 17 '12 at 14:52
  • $\begingroup$ Roger that..... (an edit has been made) $\endgroup$ – Tomarinator May 17 '12 at 14:53
  • $\begingroup$ can you tell me why is this step wrong? $\int \frac{1}{1+t^4}=ln(1+t^4)/4t^3$ $\endgroup$ – Tomarinator May 17 '12 at 14:57
  • $\begingroup$ @Saurav, try differentiating that logarithmic expression you have and see if you recover the integrand... $\endgroup$ – J. M. is a poor mathematician May 17 '12 at 14:57
  • $\begingroup$ aha got it, then i guess i will try it different way,(similar to norberts though, but a little bit different) $\endgroup$ – Tomarinator May 17 '12 at 15:00

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