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I thought this integral was simple, but it turns out it's not.

$$I=\int \frac{x}{\left(1-x^3\right)\sqrt{1-x^2}}\ dx$$

I tried the substitution $1-x^3=\frac{1}{t}$, but that leaves me again with $x = \sqrt{1-t^2}$, which doesn't simplify things. Then I tried $x = \sin(t)$, which gets me to: $$\int \frac{\sin t}{\left(1-\sin t\right)\left(1+\sin t+\sin^2 t\right)}dt$$

This looked solvable so I did another substitution: $\tan(\frac{t}{2})=u$, after a few steps I got this:

$$\int \:\frac{\left(u+u^3\right)du}{\left(u^2-2u+1\right)\left(2u^3+3u^2+4u+1\right)}$$

which is still too complicated. What am I missing? Any ideas?

Update: After splitting this function into

$$I=\frac{1}{3}\int \frac{1}{\left(1-x\right)\sqrt{1-x^2}}dx + \frac{1}{3}\int \frac{x-1}{\left(1+x+x^2\right)\sqrt{1-x^2}}dx$$

It gets a bit easier. I think I can solve the first integral. But, how do I solve the second?

Thank you!

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    $\begingroup$ After a quick check on Wolfram, this is doable, but a little nightmare. I mean an awful nightmare. $\endgroup$
    – user65203
    Commented Apr 22, 2016 at 16:46
  • $\begingroup$ please see what WA gives wolframalpha.com/input/… $\endgroup$ Commented Apr 22, 2016 at 16:48
  • $\begingroup$ Use partial fraction decomposition, $$\dfrac y{1-y^3}=\dfrac A{1-y}+\dfrac{By+C}{1+y+y^2}$$ $$\implies y=A(1+y+y^2)+(1-y)(By+C)$$ Or $y^3-1=(y-1)(y-\omega)(y-\omega^2)$ $y=1\implies3A=1$ Constant $\implies0=A+C\iff C=-A $ Coefficients of $x^2\implies 0=A-B\iff A=B$ $\endgroup$ Commented Apr 22, 2016 at 16:49
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    $\begingroup$ see math.stackexchange.com/questions/1754586 $\endgroup$
    – user84413
    Commented Apr 22, 2016 at 23:39
  • $\begingroup$ How is this duplicate question asked 30 minutes after the above linked one? $\endgroup$ Commented Apr 3, 2023 at 13:49

1 Answer 1

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After splitting the integral as follows \begin{align} &\int \frac{x}{\left(1-x^3\right)\sqrt{1-x^2}}\ dx \\ =& \ \frac{1}{3}\int \frac{1}{\left(1-x\right)\sqrt{1-x^2}}dx + \frac{1}{3}\int \frac{x-1}{\left(1+x+x^2\right)\sqrt{1-x^2}}dx \end{align} apply the Euler substitution ${\sqrt{1-x^2}}=({1+x})t$ to evaluate the two integrals

\begin{align} &\int \frac{1}{\left(1-x\right)\sqrt{1-x^2}}dx =-\int \frac1{t^2}dt=\frac1t\\ \\ &\int \frac{x-1}{\left(1+x+x^2\right)\sqrt{1-x^2}}dx\\ =&\int\frac{4t^2}{t^4+3}dt = 2\int\frac{t^2+\sqrt3}{t^4+3}+ \frac{t^2-\sqrt3}{t^4+3} \ dt\\ = &\frac{2}{\sqrt{2\sqrt3}}\tan^{-1}\frac{t^2-\sqrt3}{t\sqrt{2\sqrt3}} - \frac{2}{\sqrt{2\sqrt3}}\coth^{-1}\frac{t^2+\sqrt3}{t\sqrt{2\sqrt3}} \end{align}

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