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I am thinking of a trig sub of $x^2 = \tan{t}$ but its not leading to a nice trigonmetric form, which i can integrate. Our teacher said that it can be computed using elementary methods, but I'm unable to think of the manipualtion.

q123LsaB
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    Is there a misprint? The result isn't nice. – J.G. Aug 24 '22 at 18:30
  • I don't think wolfram could compute the integral effectively, there should be a nice answer according to our teacher, and the question is not a misprint. – q123LsaB Aug 24 '22 at 18:34
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    OK, I just thought it was weird the integrand hadn't been simplified to $\frac{\sqrt{1+x^4}}{1-x^4}$. – J.G. Aug 24 '22 at 18:39
  • FriCAS has a more complete implementation of Risch's algorithm. It gives $\frac{1}{4\sqrt{2}}\left(\log\left(\frac{4x\sqrt{1+x^4}+(x^4+2x^2+1)\sqrt{2}}{x^4-2x^2+1}\right)+2\arctan\left(\frac{2x}{\sqrt{2}\sqrt{1+x^4}}\right)\right)$. – plop Aug 24 '22 at 19:09
  • @user85667 The second term in that expression is mildly suspicious since you could also write it as $\arcsin \frac{\sqrt{2} x}{1 + x^2}$, which in turn bears a resemblance to the form of $\sin \theta$ in the Weierstrass substitution. My guess is that there's a clever trick involving multiple substitutions and a bit of partial fraction decomposition, but there's still a lot of work to make it tractable. – ConMan Aug 25 '22 at 00:39
  • @ConMan In a sense, all integrals that result in an elementary function can be solved like that, substitutions and partial fraction decomposition. Note also that $\arctan$ is really a logarithm in disguise. What is inside of these two logarithms are the substitutions. – plop Aug 25 '22 at 01:50
  • True, true. Although it's only a "logarithm in disguise" if you're allowed to peek into complex numbers. Which probably doesn't count as "elementary methods" but it sure would make things a lot easier. – ConMan Aug 25 '22 at 02:37

2 Answers2

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With parts of the calculation left as an exercise:

The surd-in-the-denominator expression I commented on is actually a subtle hint. Rewriting the numerator as$$1+x^4=\frac12(1-x^2)^2+\frac12(1+x^2)^2,$$we want to separately evaluate$$\frac12\int\frac{(1-x^2)^2}{(1-x^4)\sqrt{1+x^4}}dx=\frac12\int\frac{(1-x^2)}{(1+x^2)\sqrt{1+x^4}}dx$$and$$\frac12\int\frac{(1+x^2)^2}{(1-x^4)\sqrt{1+x^4}}dx=\frac12\int\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^4}}dx.$$Write the first part as$$\frac{1}{2\sqrt{2}}\int\frac{dy}{\sqrt{1-y^2}}=\frac{1}{2\sqrt{2}}\arcsin y+C_1$$with $y:=\frac{x\sqrt{2}}{1+x^2}$. Similarly, with $z:=\frac{x\sqrt{2}}{\sqrt{1+x^4}}$ the second part is$$\frac{1}{2\sqrt{2}}\int\frac{dz}{1-z^2}=\frac{1}{2\sqrt{2}}\operatorname{artanh}z+C_2.$$ Adding these, subsume the integration constants into one viz. $C=C_1+C_2$.

J.G.
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\begin{aligned}\int{\frac{1+x^{4}}{\left(1-x^{4}\right)\sqrt{1+x^{4}}}\,\mathrm{d}x}&=\int{\frac{1+x^{4}}{\left(1-x^{2}\right)\left(1+x^{2}\right)\sqrt{1+x^{4}}}\,\mathrm{d}x}\\ &=\int{\frac{x^{2}+\frac{1}{x^{2}}}{\left(1-x^{2}\right)\left(x+\frac{1}{x}\right)\sqrt{x^{2}+\frac{1}{x^{2}}}}\,\mathrm{d}x}\\ &=\int{\frac{x^{2}+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}\left(x+\frac{1}{x}\right)\sqrt{x^{2}+\frac{1}{x^{2}}}}\left(\frac{1}{x^{2}}-1\right)\mathrm{d}x}\\ &=\int{\frac{\left(x+\frac{1}{x}\right)^{2}-2}{\left(\left(x+\frac{1}{x}\right)^{2}-4\right)\left(x+\frac{1}{x}\right)\sqrt{\left(x+\frac{1}{x}\right)^{2}-2}}\left(1-\frac{1}{x^{2}}\right)\mathrm{d}x}\\ &=\int{\frac{y^{2}-2}{y\left(y^{2}-4\right)\sqrt{y^{2}-2}}\,\mathrm{d}y}\end{aligned}

Using another substitution $ y = \sqrt{2}\cosh{t} $. We're left with the integral : $$ \frac{1}{2}\int{\frac{\tanh{t}}{\sinh^{2}{t}-1}\,\mathrm{d}t} $$

I'll let you take it from here.

Of caurse we made the assumption that we're working in a domain $ \subset \left]0,+\infty\right[ $, so we could have $ \sqrt{x^{2}}=x $. But the same approach will stand if we're working on a domain of $ \left]-\infty, 0\right[ $.

CHAMSI
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