log(1+x)の冪級数

\(\log{\left(1+x\right)}\)の冪級数

$$\begin{eqnarray} \log{\left(1+x\right)} &=&\int_0^x \frac{1}{1+t} \mathrm{d}t \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}x}\log{x}=\frac{1}{x}より．ただしx\gt-1 \\&=&\int_0^x \frac{1+t-t}{1+t} \mathrm{d}t \;\ldots\;分子にt-tを加える \\&=&\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 \\&=&\left[x-0\right] -\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 \;\ldots\;分子にt^2-t^2を加える \\&=&x -\int_0^x \frac{t\left(1+t\right)-t^2}{1+t}\mathrm{d}t \\&=&x -\int_0^x \frac{t\left(1+t\right)}{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-\left[\frac{1}{2}x^2-\frac{1}{2}0^2\right] +\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 \;\ldots\;分子にt^3-t^3を加える \\&=&x-\frac{1}{2}x^2 +\int_0^x\frac{t^2\left(1+t\right)-t^3}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2 +\int_0^x\frac{t^2\left(1+t\right)}{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+\left[\frac{1}{3}x^3-\frac{1}{3}0^3\right] -\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 \;\ldots\;分子にt^4-t^4を加える \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3 -\int_0^x \frac{t^3\left(1+t\right)-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-\left[\frac{1}{4}x^4-\frac{1}{4}0^4\right] +\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}^{\infty} \frac{\left(-1\right)^{k+1}}{k}x^k \end{eqnarray}$$