https://spherical-harmonics.hateblo.jp/entry/Kyodai/1935/Kougaku

2025.01.23記

[1] $\displaystyle\int\dfrac{x^2}{\sqrt{1-x^2}}\,dx$ ， $\displaystyle\int\dfrac{e^x-1}{e^x+1}\,dx$

2025.01.29記

[解答]
$I=\displaystyle\int\dfrac{x^2}{\sqrt{1-x^2}}\,dx$ とおくと
$-I =\displaystyle\int x\cdot \dfrac{-x}{\sqrt{1-x^2}}\,dx$ $=x\cdot \sqrt{1-x^2}-\displaystyle\int \sqrt{1-x^2}\, dx$ $=x\sqrt{1-x^2}-\displaystyle\int \dfrac{1-x^2}{\sqrt{1-x^2}}\, dx$ $=\dfrac{1}{2}x\sqrt{1-x^2}-\mbox{Arcsin}\, x +I$
であるから，
$I=\dfrac{1}{2}\left\{\mbox{Arcsin}\, x -x\sqrt{1-x^2}\right)+(積分定数)$
である．

また，
$\displaystyle\int\dfrac{e^x-1}{e^x+1}\,dx$ $=\displaystyle\int\dfrac{e^x}{e^x+1}\,dx$ $+\displaystyle\int\dfrac{-e^{-x}}{1+e^{-x}}\,dx$ $=\log(e^x+1)+\log(e^{-x}+1)+(積分定数)$ $=\log\dfrac{(e^{x}+1)^2}{e^x}+(積分定数)$ $=2\log(e^{x}+1)-x+(積分定数)$
である．

[別解]
$\displaystyle\int\dfrac{e^x-1}{e^x+1}\,dx$ $=\displaystyle\int\dfrac{2e^x-(e^x+1)}{e^x+1}\,dx$ $=\displaystyle\int\left\{ \dfrac{2e^x}{e^x+1}-1\right\}\,dx$ $=2\log(e^{x}+1)-x+(積分定数)$
である．