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Consider the indefinite integral $$F_{\pm}(x):=\int \frac{2x^2-1}{\pm x(2x^2-1)+2x+ \sqrt{1+2x^2}} \mathbb{d} x.$$ For all $x\in \mathbb{R}$ it is $$ F_{+}(x)= \log(-x + \sqrt{1 + 2 x^2}).$$ But what is $F_{-}$?

Note that numerical studies suggest that it does exist in a certain range. Math software already struggles to prove the formula for $F_{+}$.

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  • $\begingroup$ I don't think your answer for $F_+$ is itself correct. Check wolframalpha.com/input/… With regards to $F_{-}$, that's a tough one. I'm pretty sure that you cannot solve it using any elementary function, but let's wait for other people to see. $\endgroup$
    – Fede1
    Commented Nov 17, 2020 at 20:03
  • $\begingroup$ @Fede1 It's correct, compare the expressions by using a lot of simplification. $\endgroup$
    – JHT
    Commented Nov 17, 2020 at 20:11
  • $\begingroup$ fair enough then. With regards to $F_-$, I cannot help, sorry. $\endgroup$
    – Fede1
    Commented Nov 17, 2020 at 20:14
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    $\begingroup$ Since $y = f_-(x)$ is a rational curve, $\int y dx$ is elementary (this is probably not the simplest form, I just used a CAS): $$\frac {34 - 2 \sqrt {17}} {68} \operatorname {arctanh} \omega_+ - \frac {34 + 2 \sqrt {17}} {68} \operatorname {arctanh} \omega_- - \\ \frac {\sqrt {17}} {34} \operatorname {arctanh} \frac {4 x^2 - 5} {\sqrt {17}} - \frac 1 4 \ln(2 x^4 - 5 x^2 + 1), \\ \omega_{\pm} = \frac {2 x \sqrt {4 x^2 + 2} - 4 x^2 + 5 \pm \sqrt {17}} {\sqrt {34} \pm 3 \sqrt 2}.$$ $\endgroup$
    – Maxim
    Commented Nov 18, 2020 at 12:07
  • $\begingroup$ @Maxim Nice find! Its a fairly nice formula. What CAS did you use? $\endgroup$
    – JHT
    Commented Nov 18, 2020 at 12:21

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The problem is $$F_\pm(x)=\int\frac{2x^2-1}{\pm x(2x^2-1)+2x+\sqrt{1+2x^2}}\,\mathrm{d}x.$$ Let $x=2t/(2-t^2)$, then $$\begin{aligned}F_+(t)&=2\int\bigg(\frac{t}{2-t^2}+\frac{1+t}{t^2+2t+2}\bigg)\,\mathrm{d}t,\\F_-(t)&=2\int\bigg(\frac{t}{t^2-2}-\frac{t^3-t^2-6t-2}{t^4-2t^3-12t^2-4t+4}\bigg)\,\mathrm{d}t.\end{aligned}$$ You can integrate $F_+$ directly: $$F_+(t)=-\log(2-t^2)+\log(t^2+2t+2)+C.$$ For $F_-$ you need only solve a quadratic equation to factor the denominator and then integrate directly: $$\begin{aligned}F_-(t)=\log(2-t^2)&+\frac{-17+\sqrt{17}}{34}\log(t^2+(\sqrt{17}-1)t+2)\\&+\frac{-17-\sqrt{17}}{34}\log(-t^2+(\sqrt{17}+1)t-2)+C.\end{aligned}$$ After this, solve a quadratic equation to get a formula for $F_\pm(x)$.

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  • $\begingroup$ It should be $x = 2 t/(2 - t^2)$, then $2 x^2 + 1$ becomes a complete square. Also, since you chose the plus sign when simplifying $\sqrt {2 x^2 + 1}$, you have to take the minus sign in $t = (-1 \pm \sqrt {2 x^2 + 1})/x$ when doing back-substitution. $\endgroup$
    – Maxim
    Commented Dec 2, 2020 at 23:42

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