次の3つの関数の不定積分を求めます。 (1) $\int \frac{1}{(x+1)\sqrt{1-x+x^2}} dx$ (2) $\int \sqrt{x+\sqrt{x^2+2}} dx$ (3) $\int x^3 \sqrt{1+x^2} dx$

1. 問題の内容

次の3つの関数の不定積分を求めます。

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

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

(3) $\int x^3 \sqrt{1+x^2} dx$

2. 解き方の手順

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

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

$1-x+x^2 = 1 - (\frac{1}{t} - 1) + (\frac{1}{t}-1)^2 = 1 - \frac{1}{t} + 1 + \frac{1}{t^2} - \frac{2}{t} + 1 = \frac{1}{t^2} - \frac{3}{t} + 3 = \frac{1-3t+3t^2}{t^2}$

したがって、

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

ここで、$1-3t+3t^2 = 3(t^2 - t) + 1 = 3(t^2 - t + \frac{1}{4}) - \frac{3}{4} + 1 = 3(t-\frac{1}{2})^2 + \frac{1}{4} = (\sqrt{3}(t-\frac{1}{2}))^2 + (\frac{1}{2})^2$

$u = \sqrt{3}(t-\frac{1}{2})$と置換すると、$du = \sqrt{3} dt$ より、$dt = \frac{1}{\sqrt{3}} du$

$-\int \frac{1}{\sqrt{1-3t+3t^2}} dt = -\int \frac{1}{\sqrt{u^2 + (\frac{1}{2})^2}} \frac{1}{\sqrt{3}} du = -\frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{u^2 + (\frac{1}{2})^2}} du = -\frac{1}{\sqrt{3}} \sinh^{-1}(\frac{u}{1/2}) + C$

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

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

$x = \sqrt{2} \sinh(u)$ と置換すると、$dx = \sqrt{2} \cosh(u) du$。すると、

$x^2 + 2 = 2 \sinh^2(u) + 2 = 2(\sinh^2(u) + 1) = 2\cosh^2(u)$

$\sqrt{x^2+2} = \sqrt{2}\cosh(u)$

$\sqrt{x + \sqrt{x^2+2}} = \sqrt{\sqrt{2}\sinh(u) + \sqrt{2}\cosh(u)} = \sqrt{\sqrt{2}(\sinh(u) + \cosh(u))} = \sqrt{\sqrt{2}e^u}$

$\int \sqrt{x+\sqrt{x^2+2}} dx = \int \sqrt{\sqrt{2}e^u} \sqrt{2}\cosh(u) du = \sqrt[4]{2} \sqrt{2} \int e^{u/2} \cosh(u) du = \sqrt[4]{8} \int e^{u/2} \frac{e^u + e^{-u}}{2} du = \frac{\sqrt[4]{8}}{2} \int (e^{3u/2} + e^{-u/2}) du = \frac{\sqrt[4]{8}}{2} (\frac{2}{3}e^{3u/2} - 2e^{-u/2}) + C = \frac{\sqrt[4]{8}}{3}e^{3u/2} - \sqrt[4]{8}e^{-u/2} + C$

$x = \sqrt{2} \sinh(u)$より、$u = \sinh^{-1}(\frac{x}{\sqrt{2}})$

$= \frac{\sqrt[4]{8}}{3}e^{\frac{3}{2}\sinh^{-1}(\frac{x}{\sqrt{2}})} - \sqrt[4]{8}e^{-\frac{1}{2}\sinh^{-1}(\frac{x}{\sqrt{2}})} + C$

(3) $\int x^3 \sqrt{1+x^2} dx$

$u = 1+x^2$ と置換すると、$du = 2x dx$ より、$x dx = \frac{1}{2} du$。また、$x^2 = u-1$。

$\int x^3 \sqrt{1+x^2} dx = \int x^2 \sqrt{1+x^2} x dx = \int (u-1)\sqrt{u} \frac{1}{2} du = \frac{1}{2} \int (u^{3/2} - u^{1/2}) du = \frac{1}{2} (\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}) + C = \frac{1}{5}u^{5/2} - \frac{1}{3}u^{3/2} + C = \frac{1}{5}(1+x^2)^{5/2} - \frac{1}{3}(1+x^2)^{3/2} + C$

3. 最終的な答え

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

(2) $\frac{\sqrt[4]{8}}{3}e^{\frac{3}{2}\sinh^{-1}(\frac{x}{\sqrt{2}})} - \sqrt[4]{8}e^{-\frac{1}{2}\sinh^{-1}(\frac{x}{\sqrt{2}})} + C$

(3) $\frac{1}{5}(1+x^2)^{5/2} - \frac{1}{3}(1+x^2)^{3/2} + C$