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Evaluation of $\displaystyle \int \frac{x^2}{(x^4-1)\sqrt{x^4+1}}dx$

What I try : Using Substution, $\displaystyle x=\frac{1-t}{1+t}$ and $\displaystyle dx =\frac{2}{(1+t)^2}dt$ also using $\displaystyle x^4-1=\frac{2(1+6t^2+t^4}{(1+t)^4}dt$

And $\displaystyle dx=(4t+4t^3)$

So $\displaystyle I =\frac{1}{2}\int\frac{(1-t)^2}{(1+6t^2+t^4)\sqrt{t+t^3}}dt$

Now I did not understand How I solve it after that step.

Please have a look On that , Thanks

jacky
  • 5,194
  • This integral does not have analytical solution. – NN2 Mar 02 '24 at 12:37
  • With Mathematica: $$\int \frac{x^2}{\left(x^4-1\right) \sqrt{x^4+1}} , dx=\frac{\tan ^{-1}\left(\frac{\sqrt{2} x}{\sqrt{1+x^4}}\right)-\tanh ^{-1}\left(\frac{\sqrt{2} x}{\sqrt{1+x^4}}\right)}{4 \sqrt{2}}+C$$ – Mariusz Iwaniuk Mar 02 '24 at 14:32
  • I suspect a very specific hard-to-guess trigonometric or hyperbolic substitution. – Abezhiko Mar 02 '24 at 15:16
  • @MariuszIwaniuk It's strange. I have also Mathematica and it does not give your result (but gives the same formula given by Wolfram Alpha). – NN2 Mar 02 '24 at 16:28
  • @NN2 .I use version 13.3.0 and 14.0.0 of Mathematica, give the same answers. – Mariusz Iwaniuk Mar 02 '24 at 21:18
  • @MariuszIwaniuk Mine is of version 12.0.0. Perhaps I need to upgrade it to a later version then. – NN2 Mar 02 '24 at 21:59

0 Answers0