Showing $\int_{-1}^1\frac{m(2m-1)x^{2m-2}(1-x^{2m})+m^2x^{4m-2}}{(m^2x^{4m-2}+1-x^{2m})\sqrt{1-x^{2m}}}dx=\pi$, algebraically

@user10354138' solution:

Denote the integrand by $f(x)$. We have \begin{align} &\frac{\mathrm{d}}{\mathrm{d} x}\arctan\frac{m x^{2m - 1}}{\sqrt{1 - x^{2m}}} \\ =\ & \frac{1}{1 + (\frac{m x^{2m - 1}}{\sqrt{1 - x^{2m}}})^2} \left(\frac{m(2m - 1) x^{2m - 2}}{\sqrt{1 - x^{2m}}} + \frac{m^2x^{4m - 2}}{(1 - x^{2m})\sqrt{1 - x^{2m}}} \right)\\ =\ & f(x). \end{align} Thus, $$\int_{-1}^1 f(x) \mathrm{d} x = \arctan\frac{m x^{2m - 1}}{\sqrt{1 - x^{2m}}}\Big\vert_{-1}^1 = \pi.$$

This is a supplement to @RiverLi's answer which makes @user10354138's statement that the substitution \begin{align*} \phi=\arctan\left(\frac{mx^{2m-1}}{\sqrt{1-x^{2m}}}\right)\tag{1} \end{align*} is obvious, somewhat plausible.

On the one hand we get from the definition of OPs plane curve \begin{align*} x^{2m}+y^2&=1, y\geq 0\\\\ \color{blue}{y}&\color{blue}{=\sqrt{1-x^{2m}}} \end{align*} On the other hand we have according to formula (5) the tangential angle $\phi$ of a curvature is \begin{align*} \tan\left(\phi\right)&=\frac{dy}{dx}\\ &=\frac{d}{dx}\sqrt{1-x^{2m}}\\ &\,\,\color{blue}{=-\frac{mx^{2m-1}}{\sqrt{1-x^{2m}}}} \end{align*} which makes the substitution (1) plausible.