次の不定積分を計算します。 $\int \frac{x}{(1+x)\sqrt{x^2 - x + 1}} dx$

## (7) の問題

1. 問題の内容

次の不定積分を計算します。

$\int \frac{x}{(1+x)\sqrt{x^2 - x + 1}} dx$

2. 解き方の手順

$x = \frac{1}{t}$ と置換します。すると $dx = -\frac{1}{t^2}dt$ となります。

$\int \frac{x}{(1+x)\sqrt{x^2 - x + 1}} dx = \int \frac{\frac{1}{t}}{(1+\frac{1}{t})\sqrt{\frac{1}{t^2} - \frac{1}{t} + 1}} (-\frac{1}{t^2}) dt$

$= \int \frac{\frac{1}{t}}{(\frac{t+1}{t})\sqrt{\frac{1-t+t^2}{t^2}}} (-\frac{1}{t^2}) dt = \int \frac{\frac{1}{t}}{(\frac{t+1}{t})\frac{\sqrt{1-t+t^2}}{|t|}} (-\frac{1}{t^2}) dt$

ここで、$x > 0$ 、つまり $t > 0$ のとき、$|t| = t$ となるので

$= \int \frac{1}{(\frac{t+1}{1})\frac{\sqrt{1-t+t^2}}{1}} (-\frac{1}{t^2}) \frac{1}{t} dt = \int \frac{1}{t+1} \frac{1}{\sqrt{1-t+t^2}} (-\frac{1}{t}) dt$

$= -\int \frac{1}{(t+1)\sqrt{t^2-t+1}}dt$

次に、$t+1 = \frac{1}{u}$ と置換すると、$t = \frac{1}{u} - 1 = \frac{1-u}{u}$、$dt = -\frac{1}{u^2}du$ となります。

$-\int \frac{1}{(t+1)\sqrt{t^2-t+1}}dt = -\int \frac{1}{\frac{1}{u} \sqrt{(\frac{1-u}{u})^2 - \frac{1-u}{u} + 1}} (-\frac{1}{u^2}) du$

$= \int \frac{1}{\frac{1}{u} \sqrt{\frac{1-2u+u^2 - u(1-u) + u^2}{u^2}}} (\frac{1}{u^2}) du = \int \frac{1}{\frac{1}{u} \frac{\sqrt{1-3u+3u^2}}{u}} (\frac{1}{u^2}) du$

$= \int \frac{u^2}{\sqrt{3u^2 - 3u + 1}} \frac{1}{u^2} du = \int \frac{1}{\sqrt{3u^2 - 3u + 1}} du = \frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{u^2 - u + \frac{1}{3}}}du$

$= \frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{(u - \frac{1}{2})^2 + \frac{1}{3} - \frac{1}{4}}}du = \frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{(u-\frac{1}{2})^2 + \frac{1}{12}}}du$

$w = u - \frac{1}{2}$ とすると、$dw = du$ となり

$= \frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{w^2 + \frac{1}{12}}} dw = \frac{1}{\sqrt{3}} \sinh^{-1} ( \frac{w}{\frac{1}{\sqrt{12}}} ) + C$

$= \frac{1}{\sqrt{3}} \sinh^{-1} (\sqrt{12} w) + C = \frac{1}{\sqrt{3}} \sinh^{-1} (\sqrt{12}(u-\frac{1}{2})) + C = \frac{1}{\sqrt{3}} \sinh^{-1} (\sqrt{12}(\frac{1}{t+1} - \frac{1}{2})) + C$

$= \frac{1}{\sqrt{3}} \sinh^{-1} (\sqrt{12}(\frac{1}{\frac{1}{x}+1} - \frac{1}{2})) + C = \frac{1}{\sqrt{3}} \sinh^{-1} (\sqrt{12}(\frac{x}{1+x} - \frac{1}{2})) + C$

$= \frac{1}{\sqrt{3}} \sinh^{-1} (\sqrt{12}(\frac{2x-(1+x)}{2(1+x)})) + C = \frac{1}{\sqrt{3}} \sinh^{-1} (\sqrt{12}(\frac{x-1}{2(1+x)})) + C = \frac{1}{\sqrt{3}} \sinh^{-1} ( \frac{\sqrt{3}(x-1)}{1+x}) + C$

3. 最終的な答え

$\int \frac{x}{(1+x)\sqrt{x^2 - x + 1}} dx = \frac{1}{\sqrt{3}} \sinh^{-1} (\frac{\sqrt{3}(x-1)}{x+1}) + C$

## (8) の問題

1. 問題の内容

次の不定積分を計算します。

$\int \frac{1}{(1+x)\sqrt{2+x-x^2}} dx$

2. 解き方の手順

$2+x-x^2 = -(x^2 - x - 2) = -( (x - \frac{1}{2})^2 - \frac{1}{4} - 2 ) = -( (x - \frac{1}{2})^2 - \frac{9}{4}) = \frac{9}{4} - (x-\frac{1}{2})^2$

$x+1 = \frac{3}{2} \sin(\theta) + \frac{1}{2} + 1 = \frac{3}{2} \sin(\theta) + \frac{3}{2}$

$x = \frac{1}{t}$と置換すると、$dx = -\frac{1}{t^2} dt$。

$\int \frac{1}{(1+x)\sqrt{2+x-x^2}} dx = \int \frac{1}{(1+\frac{1}{t})\sqrt{2+\frac{1}{t}-\frac{1}{t^2}}} (-\frac{1}{t^2}) dt$

$= \int \frac{1}{(\frac{t+1}{t})\sqrt{\frac{2t^2+t-1}{t^2}}} (-\frac{1}{t^2}) dt = \int \frac{1}{\frac{t+1}{t} \frac{\sqrt{2t^2+t-1}}{|t|}}(-\frac{1}{t^2}) dt = \int \frac{t^2}{(t+1)\sqrt{2t^2+t-1}} (-\frac{1}{t^2})dt = -\int \frac{1}{(t+1)\sqrt{2t^2+t-1}} dt$

3. 最終的な答え

解法が不明なため、解答できません。