3.4.58 \(\int \frac {\log (1-x^2)}{2-x^2} \, dx\) [358]

Optimal. Leaf size=239 \[ \sqrt {2} \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\frac {2 \sqrt {2}}{\sqrt {2}+x}\right )-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\frac {4 (1+x)}{\left (2+\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{\sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-x^2\right )}{\sqrt {2}}-\frac {\text {Li}_2\left (1-\frac {2 \sqrt {2}}{\sqrt {2}+x}\right )}{\sqrt {2}}+\frac {\text {Li}_2\left (1+\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{2 \sqrt {2}}+\frac {\text {Li}_2\left (1-\frac {4 (1+x)}{\left (2+\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{2 \sqrt {2}} \]

[Out]

1/2*arctanh(1/2*x*2^(1/2))*ln(-x^2+1)*2^(1/2)-1/2*arctanh(1/2*x*2^(1/2))*ln(-4*(1-x)/(2-2^(1/2))/(x+2^(1/2)))*
2^(1/2)-1/2*arctanh(1/2*x*2^(1/2))*ln(4*(1+x)/(2+2^(1/2))/(x+2^(1/2)))*2^(1/2)+1/4*polylog(2,1+4*(1-x)/(2-2^(1
/2))/(x+2^(1/2)))*2^(1/2)-1/2*polylog(2,1-2*2^(1/2)/(x+2^(1/2)))*2^(1/2)+1/4*polylog(2,1-4*(1+x)/(2+2^(1/2))/(
x+2^(1/2)))*2^(1/2)+arctanh(1/2*x*2^(1/2))*ln(2*2^(1/2)/(x+2^(1/2)))*2^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {212, 2520, 12, 6139, 6057, 2449, 2352, 2497} \begin {gather*} -\frac {\text {PolyLog}\left (2,1-\frac {2 \sqrt {2}}{x+\sqrt {2}}\right )}{\sqrt {2}}+\frac {\text {PolyLog}\left (2,\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}+1\right )}{2 \sqrt {2}}+\frac {\text {PolyLog}\left (2,1-\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right )}{2 \sqrt {2}}+\frac {\log \left (1-x^2\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}+\sqrt {2} \log \left (\frac {2 \sqrt {2}}{x+\sqrt {2}}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {\log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\log \left (\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[1 - x^2]/(2 - x^2),x]

[Out]

Sqrt[2]*ArcTanh[x/Sqrt[2]]*Log[(2*Sqrt[2])/(Sqrt[2] + x)] - (ArcTanh[x/Sqrt[2]]*Log[(-4*(1 - x))/((2 - Sqrt[2]
)*(Sqrt[2] + x))])/Sqrt[2] - (ArcTanh[x/Sqrt[2]]*Log[(4*(1 + x))/((2 + Sqrt[2])*(Sqrt[2] + x))])/Sqrt[2] + (Ar
cTanh[x/Sqrt[2]]*Log[1 - x^2])/Sqrt[2] - PolyLog[2, 1 - (2*Sqrt[2])/(Sqrt[2] + x)]/Sqrt[2] + PolyLog[2, 1 + (4
*(1 - x))/((2 - Sqrt[2])*(Sqrt[2] + x))]/(2*Sqrt[2]) + PolyLog[2, 1 - (4*(1 + x))/((2 + Sqrt[2])*(Sqrt[2] + x)
)]/(2*Sqrt[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 2520

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[u*(x^(n - 1)/(d + e*x^n)
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

Rule 6057

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x]))*(Log[2/
(1 + c*x)]/e), x] + (Dist[b*(c/e), Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[b*(c/e), Int[Log[2*c*((d
+ e*x)/((c*d + e)*(1 + c*x)))]/(1 - c^2*x^2), x], x] + Simp[(a + b*ArcTanh[c*x])*(Log[2*c*((d + e*x)/((c*d + e
)*(1 + c*x)))]/e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 6139

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[a
+ b*ArcTanh[c*x], x^m/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[m] &&  !(EqQ[m, 1] && NeQ[
a, 0])

Rubi steps

\begin {align*} \int \frac {\log \left (1-x^2\right )}{2-x^2} \, dx &=\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-x^2\right )}{\sqrt {2}}+2 \int \frac {x \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2} \left (1-x^2\right )} \, dx\\ &=\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-x^2\right )}{\sqrt {2}}+\sqrt {2} \int \frac {x \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{1-x^2} \, dx\\ &=\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-x^2\right )}{\sqrt {2}}+\sqrt {2} \int \left (-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{2 (-1+x)}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{2 (1+x)}\right ) \, dx\\ &=\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-x^2\right )}{\sqrt {2}}-\frac {\int \frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{-1+x} \, dx}{\sqrt {2}}-\frac {\int \frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{1+x} \, dx}{\sqrt {2}}\\ &=\sqrt {2} \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\frac {2 \sqrt {2}}{\sqrt {2}+x}\right )-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\frac {4 (1+x)}{\left (2+\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{\sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-x^2\right )}{\sqrt {2}}-2 \left (\frac {1}{2} \int \frac {\log \left (\frac {2}{1+\frac {x}{\sqrt {2}}}\right )}{1-\frac {x^2}{2}} \, dx\right )+\frac {1}{2} \int \frac {\log \left (\frac {\sqrt {2} (-1+x)}{\left (1-\frac {1}{\sqrt {2}}\right ) \left (1+\frac {x}{\sqrt {2}}\right )}\right )}{1-\frac {x^2}{2}} \, dx+\frac {1}{2} \int \frac {\log \left (\frac {\sqrt {2} (1+x)}{\left (1+\frac {1}{\sqrt {2}}\right ) \left (1+\frac {x}{\sqrt {2}}\right )}\right )}{1-\frac {x^2}{2}} \, dx\\ &=\sqrt {2} \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\frac {2 \sqrt {2}}{\sqrt {2}+x}\right )-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\frac {4 (1+x)}{\left (2+\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{\sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-x^2\right )}{\sqrt {2}}+\frac {\text {Li}_2\left (1+\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{2 \sqrt {2}}+\frac {\text {Li}_2\left (1-\frac {4 (1+x)}{\left (2+\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{2 \sqrt {2}}-2 \frac {\text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+\frac {x}{\sqrt {2}}}\right )}{\sqrt {2}}\\ &=\sqrt {2} \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\frac {2 \sqrt {2}}{\sqrt {2}+x}\right )-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\frac {4 (1+x)}{\left (2+\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{\sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-x^2\right )}{\sqrt {2}}-\frac {\text {Li}_2\left (1-\frac {2 \sqrt {2}}{\sqrt {2}+x}\right )}{\sqrt {2}}+\frac {\text {Li}_2\left (1+\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{2 \sqrt {2}}+\frac {\text {Li}_2\left (1-\frac {4 (1+x)}{\left (2+\sqrt {2}\right ) \left (\sqrt {2}+x\right )}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 248, normalized size = 1.04 \begin {gather*} -\frac {\left (\log \left (\sqrt {2}-x\right )-\log \left (\sqrt {2}+x\right )\right ) \left (-\log (-1+x)-\log (1+x)+\log \left (1-x^2\right )\right )}{2 \sqrt {2}}+\frac {\log \left (1-\frac {-1+x}{-1-\sqrt {2}}\right ) \log (-1+x)+\text {Li}_2\left (\frac {-1+x}{-1-\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\log \left (1-\frac {-1+x}{-1+\sqrt {2}}\right ) \log (-1+x)+\text {Li}_2\left (\frac {-1+x}{-1+\sqrt {2}}\right )}{2 \sqrt {2}}+\frac {\log (1+x) \log \left (1-\frac {1+x}{1-\sqrt {2}}\right )+\text {Li}_2\left (\frac {1+x}{1-\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\log (1+x) \log \left (1-\frac {1+x}{1+\sqrt {2}}\right )+\text {Li}_2\left (\frac {1+x}{1+\sqrt {2}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[1 - x^2]/(2 - x^2),x]

[Out]

-1/2*((Log[Sqrt[2] - x] - Log[Sqrt[2] + x])*(-Log[-1 + x] - Log[1 + x] + Log[1 - x^2]))/Sqrt[2] + (Log[1 - (-1
+ x)/(-1 - Sqrt[2])]*Log[-1 + x] + PolyLog[2, (-1 + x)/(-1 - Sqrt[2])])/(2*Sqrt[2]) - (Log[1 - (-1 + x)/(-1 +
Sqrt[2])]*Log[-1 + x] + PolyLog[2, (-1 + x)/(-1 + Sqrt[2])])/(2*Sqrt[2]) + (Log[1 + x]*Log[1 - (1 + x)/(1 - S
qrt[2])] + PolyLog[2, (1 + x)/(1 - Sqrt[2])])/(2*Sqrt[2]) - (Log[1 + x]*Log[1 - (1 + x)/(1 + Sqrt[2])] + PolyL
og[2, (1 + x)/(1 + Sqrt[2])])/(2*Sqrt[2])

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Maple [A]
time = 0.52, size = 194, normalized size = 0.81

method	result	size
default	\(-\frac {\left (\ln \left (x -\sqrt {2}\right ) \ln \left (-x^{2}+1\right )-\dilog \left (\frac {x +1}{1+\sqrt {2}}\right )-\ln \left (x -\sqrt {2}\right ) \ln \left (\frac {x +1}{1+\sqrt {2}}\right )-\dilog \left (\frac {-1+x}{\sqrt {2}-1}\right )-\ln \left (x -\sqrt {2}\right ) \ln \left (\frac {-1+x}{\sqrt {2}-1}\right )\right ) \sqrt {2}}{4}+\frac {\left (\ln \left (x +\sqrt {2}\right ) \ln \left (-x^{2}+1\right )-\dilog \left (\frac {x +1}{-\sqrt {2}+1}\right )-\ln \left (x +\sqrt {2}\right ) \ln \left (\frac {x +1}{-\sqrt {2}+1}\right )-\dilog \left (\frac {-1+x}{-1-\sqrt {2}}\right )-\ln \left (x +\sqrt {2}\right ) \ln \left (\frac {-1+x}{-1-\sqrt {2}}\right )\right ) \sqrt {2}}{4}\)	\(194\)
risch	\(-\frac {\sqrt {2}\, \ln \left (-x^{2}+1\right ) \ln \left (x -\sqrt {2}\right )}{4}+\frac {\sqrt {2}\, \ln \left (x -\sqrt {2}\right ) \ln \left (\frac {x +1}{1+\sqrt {2}}\right )}{4}+\frac {\sqrt {2}\, \ln \left (x -\sqrt {2}\right ) \ln \left (\frac {-1+x}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \dilog \left (\frac {x +1}{1+\sqrt {2}}\right )}{4}+\frac {\sqrt {2}\, \dilog \left (\frac {-1+x}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \ln \left (-x^{2}+1\right ) \ln \left (x +\sqrt {2}\right )}{4}-\frac {\sqrt {2}\, \ln \left (x +\sqrt {2}\right ) \ln \left (\frac {x +1}{-\sqrt {2}+1}\right )}{4}-\frac {\sqrt {2}\, \ln \left (x +\sqrt {2}\right ) \ln \left (\frac {-1+x}{-1-\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \dilog \left (\frac {x +1}{-\sqrt {2}+1}\right )}{4}-\frac {\sqrt {2}\, \dilog \left (\frac {-1+x}{-1-\sqrt {2}}\right )}{4}\)	\(214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(-x^2+1)/(-x^2+2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(ln(x-2^(1/2))*ln(-x^2+1)-dilog((x+1)/(1+2^(1/2)))-ln(x-2^(1/2))*ln((x+1)/(1+2^(1/2)))-dilog((-1+x)/(2^(1
/2)-1))-ln(x-2^(1/2))*ln((-1+x)/(2^(1/2)-1)))*2^(1/2)+1/4*(ln(x+2^(1/2))*ln(-x^2+1)-dilog((x+1)/(-2^(1/2)+1))-
ln(x+2^(1/2))*ln((x+1)/(-2^(1/2)+1))-dilog((-1+x)/(-1-2^(1/2)))-ln(x+2^(1/2))*ln((-1+x)/(-1-2^(1/2))))*2^(1/2)

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Maxima [A]
time = 0.51, size = 208, normalized size = 0.87 \begin {gather*} \frac {1}{4} \, \sqrt {2} {\left ({\left (\log \left (2 \, x + 2 \, \sqrt {2}\right ) - \log \left (2 \, x - 2 \, \sqrt {2}\right )\right )} \log \left (-x^{2} + 1\right ) - \log \left (x + \sqrt {2}\right ) \log \left (-\frac {x + \sqrt {2}}{\sqrt {2} + 1} + 1\right ) + \log \left (x - \sqrt {2}\right ) \log \left (\frac {x - \sqrt {2}}{\sqrt {2} + 1} + 1\right ) - \log \left (x + \sqrt {2}\right ) \log \left (-\frac {x + \sqrt {2}}{\sqrt {2} - 1} + 1\right ) + \log \left (x - \sqrt {2}\right ) \log \left (\frac {x - \sqrt {2}}{\sqrt {2} - 1} + 1\right ) - {\rm Li}_2\left (\frac {x + \sqrt {2}}{\sqrt {2} + 1}\right ) + {\rm Li}_2\left (-\frac {x - \sqrt {2}}{\sqrt {2} + 1}\right ) - {\rm Li}_2\left (\frac {x + \sqrt {2}}{\sqrt {2} - 1}\right ) + {\rm Li}_2\left (-\frac {x - \sqrt {2}}{\sqrt {2} - 1}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-x^2+1)/(-x^2+2),x, algorithm="maxima")

[Out]

1/4*sqrt(2)*((log(2*x + 2*sqrt(2)) - log(2*x - 2*sqrt(2)))*log(-x^2 + 1) - log(x + sqrt(2))*log(-(x + sqrt(2))
/(sqrt(2) + 1) + 1) + log(x - sqrt(2))*log((x - sqrt(2))/(sqrt(2) + 1) + 1) - log(x + sqrt(2))*log(-(x + sqrt(
2))/(sqrt(2) - 1) + 1) + log(x - sqrt(2))*log((x - sqrt(2))/(sqrt(2) - 1) + 1) - dilog((x + sqrt(2))/(sqrt(2)
+ 1)) + dilog(-(x - sqrt(2))/(sqrt(2) + 1)) - dilog((x + sqrt(2))/(sqrt(2) - 1)) + dilog(-(x - sqrt(2))/(sqrt(
2) - 1)))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-x^2+1)/(-x^2+2),x, algorithm="fricas")

[Out]

integral(-log(-x^2 + 1)/(x^2 - 2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\log {\left (1 - x^{2} \right )}}{x^{2} - 2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(-x**2+1)/(-x**2+2),x)

[Out]

-Integral(log(1 - x**2)/(x**2 - 2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-x^2+1)/(-x^2+2),x, algorithm="giac")

[Out]

integrate(-log(-x^2 + 1)/(x^2 - 2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} -\int \frac {\ln \left (1-x^2\right )}{x^2-2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-log(1 - x^2)/(x^2 - 2),x)

[Out]

-int(log(1 - x^2)/(x^2 - 2), x)

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