本帖最后由 青青子衿 于 2021-3-13 09:35 编辑
\begin{align*}
\int_{0}^{x}\frac{1}{\sqrt{\left(1-a^{2}t^{2}\right)\left(1-b^{2}t^{2}\right)\left(1-c^{2}t^{2}\right)}}\mathrm{d}t\quad\qquad\\
\\
=
\frac{1}{\sqrt{c^{2}-a^{2}}}\int_{0}^{x\cdot\sqrt{\frac{c^2-a^2}{1-a^2x^2}}}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-\frac{b^{2}-a^{2}}{c^{2}-a^{2}}t^{2}\right)}}\mathrm{d}t
\end{align*}
\begin{align*}
g'(x)^2 &=4g^3(x)-4(a^2+b^2+c^2)g^2(x)+4(a^2b^2+a^2c^2+b^2c^2)g(x)-4a^2b^2c^2\\
g'(x)^2 &= 4\Big[g(x)-\alpha\,\Big]^3 - g_2 \Big[g(x)-\alpha\,\Big] - g_3\\
g'(x)^2 &= 4g^3(x)-12\alpha\,g^2(x) + (12\alpha^2-g_2)g(x) - (g_3-g_2\alpha+4\alpha^3)\\
\alpha&=\dfrac{1}{3}\left(a^2+b^2+c^2\right)\\
g_2&=\dfrac{4}{3}(a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2)\\
g_3&=\dfrac{4}{27} \left(a^2+b^2-2 c^2\right) \left(2 a^2-b^2-c^2\right) \left(a^2-2 b^2+c^2\right)
\end{align*}- f\left(x\right)=\int_{0}^{x}\frac{1}{\sqrt{\left(u-a\right)\left(u-b\right)\left(u-c\right)\left(u-d\right)}}du
- g\left(x\right)=\int_{-\frac{\beta}{\alpha}}^{-\frac{\delta x-\beta}{\gamma x-\alpha}}\frac{\alpha\delta-\beta\gamma}{\sqrt{\left(\left(\alpha-a\gamma\right)t+\beta-a\delta\right)\left(\left(\alpha-b\gamma\right)t+\beta-b\delta\right)\left(\left(\alpha-c\gamma\right)t+\beta-c\delta\right)\left(\left(\alpha-d\gamma\right)t+\beta-d\delta\right)}}dt
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发表于 2021-1-28 18:34
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