積分解説9

\begin{eqnarray} \int_{0}^{2}\frac{\mathrm{d}x}{\sqrt{1+x^3}} &=& \frac{2}{3}\int_{1}^{3}(x^2-1)^{-\frac{2}{3}}\mathrm{d}x \quad(\sqrt{1+x^3}\mapsto x) \\ &=& \frac{2^{\frac{2}{3}}}{3}\int_{0}^{1}\big(x(x+1)\big)^{-\frac{2}{3}}\mathrm{d}x \quad(\frac{x-1}{2}\mapsto x) \\ &=& \frac{2^{\frac{2}{3}}}{3}\int_{0}^{\frac{1}{2}}\big(x(1-x)\big)^{-\frac{2}{3}}\mathrm{d}x \quad (x\mapsto\frac{x}{1-x}) \\ &=& \frac{1}{2^{\frac{1}{3}}3}\int_{0}^{1}\big(x(1-x)\big)^{-\frac{2}{3}}\mathrm{d}x \quad (\because \text{被積分関数の対称性}) \\ &=& \frac{1}{2^{\frac{1}{3}}3}B\Big(\frac{1}{3},\frac{1}{3}\Big) \\ &=& \frac{1}{2^{\frac{1}{3}}3}\frac{\Gamma\big(\frac{1}{3}\big)\Gamma\big(\frac{1}{3}\big)}{\Gamma\big(\frac{2}{3}\big)} \\ &=& \frac{\Gamma^3\big(\frac{1}{3}\big)}{2^{4/3}3^{1/2}\pi} \quad(\because \text{Gamma関数の相反公式}) \end{eqnarray}