log(1+x)の冪級数

$\log (1+x)$の冪級数

$$\begin{array}{rcl}\log (1+x)& =& {\int }_0^x\frac{1}{1+t}\mathrm{d}t\dots \frac{\mathrm{d}}{\mathrm{d}x}\log x=\frac{1}{x}\mathrm{よ}\mathrm{り}．\mathrm{た}\mathrm{だ}\mathrm{し}x>-1\\ & =& {\int }_0^x\frac{1+t-t}{1+t}\mathrm{d}t\dots \mathrm{分}\mathrm{子}\mathrm{に}t-t\mathrm{を}\mathrm{加}\mathrm{え}\mathrm{る}\\ & =& {\int }_0^x\frac{1+t}{1+t}+\frac{-t}{1+t}\mathrm{d}t\\ & =& {\int }_0^{x}1+\frac{-t}{1+t}\mathrm{d}t\\ & =& {\int }_0^{x}1\mathrm{d}t-{\int }_0^x\frac{t}{1+t}\mathrm{d}t\\ & =& {\left[t\right]}_0^x-{\int }_0^x\frac{t}{1+t}\mathrm{d}t\\ & =& [x-0]-{\int }_0^x\frac{t}{1+t}\mathrm{d}t\\ & =& x-{\int }_0^x\frac{t}{1+t}\mathrm{d}t\\ & =& x-{\int }_0^x\frac{t+t^2-t^2}{1+t}\mathrm{d}t\dots \mathrm{分}\mathrm{子}\mathrm{に}{t}^2-t^2\mathrm{を}\mathrm{加}\mathrm{え}\mathrm{る}\\ & =& x-{\int }_0^x\frac{t(1+t)-t^2}{1+t}\mathrm{d}t\\ & =& x-{\int }_0^x\frac{t(1+t)}{1+t}+\frac{-t^2}{1+t}\mathrm{d}t\\ & =& x-{\int }_0^{x}t+\frac{-t^2}{1+t}\mathrm{d}t\\ & =& x-{\int }_0^{x}t\mathrm{d}t-{\int }_0^x\frac{-t^2}{1+t}\mathrm{d}t\\ & =& x-{\int }_0^{x}t\mathrm{d}t+{\int }_0^x\frac{t^2}{1+t}\mathrm{d}t\\ & =& x-{\left[\frac{1}{2}{t}^2\right]}_0^x+{\int }_0^x\frac{t^2}{1+t}\mathrm{d}t\\ & =& x-[\frac{1}{2}{x}^2-\frac{1}{2}{0}^2]+{\int }_0^x\frac{t^2}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+{\int }_0^x\frac{t^2}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+{\int }_0^x\frac{t^2+t^3-t^3}{1+t}\mathrm{d}t\dots \mathrm{分}\mathrm{子}\mathrm{に}{t}^3-t^3\mathrm{を}\mathrm{加}\mathrm{え}\mathrm{る}\\ & =& x-\frac{1}{2}{x}^2+{\int }_0^x\frac{t^2(1+t)-t^3}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+{\int }_0^x\frac{t^2(1+t)}{1+t}\frac{-t^3}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+{\int }_0^x{t}^2+\frac{-t^3}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+{\int }_0^x{t}^2\mathrm{d}t-{\int }_0^x\frac{t^3}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+{\left[\frac{1}{3}{t}^3\right]}_0^x-{\int }_0^x\frac{t^3}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+[\frac{1}{3}{x}^3-\frac{1}{3}{0}^3]-{\int }_0^x\frac{t^3}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-{\int }_0^x\frac{t^3}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-{\int }_0^x\frac{t^3+t^4-t^4}{1+t}\mathrm{d}t\dots \mathrm{分}\mathrm{子}\mathrm{に}{t}^4-t^4\mathrm{を}\mathrm{加}\mathrm{え}\mathrm{る}\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-{\int }_0^x\frac{t^3(1+t)-t^4}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-{\int }_0^x{t}^3+\frac{-t^4}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-{\int }_0^x{t}^3\mathrm{d}t-{\int }_0^x\frac{-t^4}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-{\int }_0^x{t}^3\mathrm{d}t+{\int }_0^x\frac{t^4}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-{\left[\frac{1}{4}{t}^4\right]}_0^x+{\int }_0^x\frac{t^4}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-[\frac{1}{4}{x}^4-\frac{1}{4}{0}^4]+{\int }_0^x\frac{t^4}{1+t}\mathrm{d}t\\ & =& x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+{\int }_0^x\frac{t^4}{1+t}\mathrm{d}t\\ & =& \sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k+1}}{k}{x}^k\end{array}$$