Numerical methods surely indicate that $\int_0^{\frac{1}{3}} 2 \sqrt{9 x+1} \sqrt{21 x-4 \sqrt{3} \sqrt{x (9 x+1)}+1} \left(4 \sqrt{3} \sqrt{x (9 x+1)}+1\right) \, dx= \frac{7}{9}$.
Can this be formally demonstrated?
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Sign up to join this communityNumerical methods surely indicate that $\int_0^{\frac{1}{3}} 2 \sqrt{9 x+1} \sqrt{21 x-4 \sqrt{3} \sqrt{x (9 x+1)}+1} \left(4 \sqrt{3} \sqrt{x (9 x+1)}+1\right) \, dx= \frac{7}{9}$.
Can this be formally demonstrated?
Yes it can, because the long square root equals $\sqrt{9x+1}-\sqrt{12 x}$. After this observation you get a standard integral, which reduces to an integral of rational function if you change the variable to $\sqrt{9/x+1}$.
Well, pursuing (as best I can) the line of reasoning put forth by K B Dave in his (stereographic-projection-motivated) answer to the related (antecedent) question in https://math.stackexchange.com/questions/3327016/can-knowledge-of-int-fx2-dx-possibly-be-used-in-obtaining-int-fx-dx, I make the transformation $x\to \frac{4 m^2}{3 \left(m^2-3\right)^2}$, having a jacobian of $\frac{8 m \left(m^2+3\right)}{3 \left(m^2-3\right)^3}$ and new limits of integration (0,1}, whereupon Mathematica directly yields the desired answer of $\frac{7}{9}$.
K B Dave had written:
Let \begin{equation} 3 x=u^2, 9 x+1=v^2, \end{equation} and stereographically project the curve $v^2-3 u^2=1$ about $(u,v) =(0,1)$, so that \begin{equation} \frac{v-1}{u}= \frac{3 u}{v+1}=m. \end{equation}
Then the integrands reduce to rational expression in $m$.,…