https://spherical-harmonics.hateblo.jp/entry/Todai/1934/Kougaku

2022.08.10記

[1] 下ノ積分ヲ求メヨ．

(a) $\displaystyle\int\dfrac{3x+4}{\sqrt{x^2+2x+5}}dx$

(b) $\displaystyle\int\dfrac{dx}{x\sqrt{1+x^6}}$

2022.08.11記

[解答]
(a) $\displaystyle\int\dfrac{3x+4}{\sqrt{x^2+2x+5}}dx$ $=\displaystyle\int\dfrac{3(x+1)+1}{\sqrt{x^2+2x+5}}dx$
$=3\sqrt{x^2+2x+5}+\displaystyle\int\dfrac{1}{\sqrt{(x+1)^2+4}}dx$
$=3\sqrt{x^2+2x+5}$ $+\log|x+1+\sqrt{x^2+2x+5}|+(積分定数)$

(b) $\sqrt{1+x^6}=t$ と置換すると $3x^5\,dx=t\,dt$ から
$\displaystyle\int\dfrac{dx}{x\sqrt{1+x^6}}$ $=\displaystyle\int\dfrac{3x^5\,dx}{3x^6\sqrt{1+x^6}}$ $=\displaystyle\int\dfrac{t\,dt}{3(t^2-1)t}$ $=\displaystyle\int\dfrac{dt}{3(t^2-1)}$ $=\dfrac{1}{6}\log\left|\dfrac{t-1}{t+1}\right|+(積分定数)$
$=\dfrac{1}{6}\log\left|\dfrac{\sqrt{1+x^6}-1}{\sqrt{1+x^6}+1}\right|+(積分定数)$