積分解説9

$$\begin{array}{rcl}{\int }_0^2\frac{\mathrm{d}x}{\sqrt{1+x^3}}& =& \frac{2}{3}{\int }_1^3(x^2-1{)}^{-\frac{2}{3}}\mathrm{d}x(\sqrt{1+x^3}\mapsto x)\\ & =& \frac{2^{\frac{2}{3}}}{3}{\int }_0^1(x(x+1){)}^{-\frac{2}{3}}\mathrm{d}x(\frac{x-1}{2}\mapsto x)\\ & =& \frac{2^{\frac{2}{3}}}{3}{\int }_0^{\frac{1}{2}}(x(1-x){)}^{-\frac{2}{3}}\mathrm{d}x(x\mapsto \frac{x}{1-x})\\ & =& \frac{1}{2^{\frac{1}{3}}3}{\int }_0^1(x(1-x){)}^{-\frac{2}{3}}\mathrm{d}x(\because \text{被積分関数の対称性})\\ & =& \frac{1}{2^{\frac{1}{3}}3}B(\frac{1}{3},\frac{1}{3})\\ & =& \frac{1}{2^{\frac{1}{3}}3}\frac{\mathrm{\Gamma }(\frac{1}{3})\mathrm{\Gamma }(\frac{1}{3})}{\mathrm{\Gamma }(\frac{2}{3})}\\ & =& \frac{{\mathrm{\Gamma }}^3(\frac{1}{3})}{2^{4/3}{3}^{1/2}\pi }(\because \text{Gamma関数の相反公式})\end{array}$$