【解説】東大理系数学2019 第一問

$$\begin{aligned} &\left( x^2 + \dfrac{x}{\sqrt{1+x^2}} \right) \left( 1 + \dfrac{x}{(1+x^2) \sqrt{1+x^2}} \right)\\ &= x^2 + \dfrac{x}{\sqrt{1+x^2}} + \dfrac{x^3}{(1+x^2) \sqrt{1+x^2}} + \dfrac{x^2}{(1+x^2)^2} \end{aligned}$$ となる。各項で計算をする。

• 1項目 $$\int_0^1 x^2 dx = \dfrac{1}{3}$$

• 2項目 $$\begin{aligned} \int_0^1 \dfrac{x}{\sqrt{1+x^2}} dx &= \int_0^1 \dfrac{1}{2} \dfrac{(1+x^2)'}{\sqrt{1+x^2}}\\ &= \Big[ (1+x^2)^{\frac{1}{2}} \Big]_0^1\\ &= \sqrt{2} - 1 \end{aligned}$$

• 3項目 $$\begin{aligned} &\int_0^1 \dfrac{x^3}{(1+x^2) \sqrt{1+x^2}} dx\\ &= \int_0^1 \left( 1 - \dfrac{1}{1+x^2} \right) \dfrac{x}{\sqrt{1+x^2}} dx\\ &= \int_0^1 \dfrac{1}{2} \left( (1+x^2)^{-\frac{1}{2}} - (1+x^2)^{-\frac{3}{2}} \right) (1+x^2)' dx\\ &= \Big[ (1+x^2)^{\frac{1}{2}} + (1+x^2)^{-\frac{1}{2}} \Big]_0^1\\ &= \sqrt{2} + \dfrac{1}{\sqrt{2}} - 1 - 1\\ &= \dfrac{3}{2} \sqrt{2} - 2 \end{aligned}$$

• 4項目 $$\begin{aligned} &\int_0^1 \dfrac{x^2}{(1+x^2)^2} dx\\ &= \dfrac{1}{2} \int_0^1 \dfrac{x(1+x^2)'}{(1+x^2)^2} dx\\ &= \dfrac{1}{2} \Big[ -\dfrac{x}{(1+x^2)} \Big]_0^1 + \dfrac{1}{2} \int_0^1 \dfrac{dx}{1+x^2} &(\text{部分積分})\\ &= -\dfrac{1}{4} + \dfrac{\pi}{8} \end{aligned}$$ なお， $$\begin{aligned} \int_0^1 \dfrac{dx}{1+x^2} &= \int_0^{\frac{\pi}{4}} \dfrac{1}{1+\tan^2 \theta} \dfrac{1}{\cos^2 \theta} d\theta\\ &= \int_0^{\frac{\pi}{4}} d\theta\\ &= \dfrac{\pi}{4} \end{aligned}$$ と計算した。

以上を足すと $$\begin{aligned} &\int_0^1 \left( x^2 + \dfrac{x}{\sqrt{1+x^2}} \right) \left( 1 + \dfrac{x}{(1+x^2) \sqrt{1+x^2}} \right) dx\\ &= \dfrac{1}{3} + (\sqrt{2} - 1) + \left( \dfrac{3}{2} \sqrt{2} - 2 \right) + \left( \dfrac{\pi}{8} - \dfrac{1}{4} \right)\\ &= -\dfrac{35}{12} + \dfrac{5}{2} \sqrt{2} + \dfrac{\pi}{8} \end{aligned}$$ となる。

$$\begin{aligned} &\int_0^1 \dfrac{x^2}{(1+x^2)^2} dx\\ &= \int_0^{\frac{\pi}{4}} \dfrac{\tan^2 \theta}{\left( \frac{1}{\cos^2 \theta} \right)^2} \dfrac{d\theta}{\cos^2 \theta}\\ &= \int_0^{\frac{\pi}{4}} \sin^2 \theta\\ &= \int_0^{\frac{\pi}{4}} \dfrac{1 - \cos 2\theta}{2} d\theta\\ &= \dfrac{1}{2} \Big[ \theta - \dfrac{1}{2} \sin 2\theta \Big]_0^{\frac{\pi}{4}}\\ &= \dfrac{\pi}{8} - \dfrac{1}{4} \end{aligned}$$

$$\begin{aligned} &\left( x^2 + \dfrac{x}{\sqrt{1+x^2}} \right) \left( 1 + \dfrac{x}{(1+x^2) \sqrt{1+x^2}} \right)\\ &= x^2 + \dfrac{x}{\sqrt{1+x^2}} + \dfrac{x^3}{(1+x^2) \sqrt{1+x^2}} + \dfrac{x^2}{(1+x^2)^2} \end{aligned}$$ と展開された。2項目と3項目について $$\begin{aligned} &\dfrac{x}{\sqrt{1+x^2}} + \dfrac{x^3}{(1+x^2) \sqrt{1+x^2}}\\ &= \dfrac{x}{\sqrt{1+x^2}} + \dfrac{x^3 + x - x}{(1+x^2) \sqrt{1+x^2}}\\ &= \dfrac{2x}{\sqrt{x^2+1}} - \dfrac{x}{(1+x^2)\sqrt{1+x^2}} \end{aligned}$$ と変形できる。

• 2項目 $$\begin{aligned} \int_0^{1} \dfrac{2x}{\sqrt{x^2+1}} dx &= \int_0^1 \dfrac{2(x^2+1)'}{\sqrt{x^2+1}} dx\\ &= \Big[ 2\sqrt{x^2+1} \Big]_0^1\\ &= 2\sqrt{2} - 2 \end{aligned}$$

• 3項目 $$\begin{aligned} \int_0^1 \dfrac{x}{(1+x^2)\sqrt{1+x^2}} dx &= \int_0^1 \dfrac{(1+x^2)'}{2(1+x^2)^{\frac{3}{2}}} dx\\ &= \Big[ -(1+x^2)^{-\frac{1}{2}} \Big]_0^1\\ &= - \dfrac{\sqrt{2}}{2} + 1 \end{aligned}$$

$$\begin{aligned} &\int_0^1 \left( x^2 + \dfrac{x}{\sqrt{1+x^2}} \right) \left( 1 + \dfrac{x}{(1+x^2) \sqrt{1+x^2}} \right) dx\\ &= \dfrac{1}{3} + (2\sqrt{2} - 2) + \left( \dfrac{\sqrt{2}}{2} - 1 \right) + \left( \dfrac{\pi}{8} - \dfrac{1}{4} \right)\\ &= -\dfrac{35}{12} + \dfrac{5}{2} \sqrt{2} + \dfrac{\pi}{8} \end{aligned}$$ となる。

$$\begin{aligned} & \int_{0}^{1} \left( x^2 + \dfrac{x}{\sqrt{1+x^2}} \right) \left( 1+\dfrac{x}{(1+x^2) \sqrt{1+x^2}} \right) dx\\ &= \int_{0}^{1} x \left(x + \dfrac{1}{\sqrt{1+x^2}} \right) \left( -x- \dfrac{1}{\sqrt{1+x^2}} \right)'dx\\ & \quad +2 \int_{0}^{1}x\left( x + \dfrac{1}{\sqrt{1+x^2}} \right)dx \end{aligned}$$

• 1項目 $$\begin{aligned} & \int_{0}^{1} x \left( x + \dfrac{1}{\sqrt{1+x^2}} \right) \left( - x - \dfrac{1}{\sqrt{1+x^2}} \right)' dx\\ &= \left[ -\dfrac{x}{2} \left( x + \dfrac{1}{\sqrt{1+x^2}} \right)^2 \right]_{0}^{1}+ \dfrac{1}{2} \int_{0}^{1} \left( x + \dfrac{1}{\sqrt{1+x^2}} \right)^2 dx\\ &= -\dfrac{1}{2} \left( 1 + \dfrac{1}{\sqrt{2}} \right)^2\\ & \quad + \dfrac{1}{2} \int_{0}^{1} \left( x^2 + \dfrac{2x}{\sqrt{1+x^2}} + \dfrac{1}{1+x^2} \right)^2 dx\\ &= -\dfrac{3}{4} - \dfrac{\sqrt{2}}{2} + \dfrac{1}{2} \left[ \dfrac{x^3}{3} + 2 \sqrt{1+x^2} \right]_{0}^{1} + \dfrac{1}{2} \int_{0}^{1} \dfrac{1}{1+x^2} dx\\ &= -\dfrac{13}{12} + \dfrac{\sqrt{2}}{2} + \dfrac{\pi}{8} \end{aligned}$$

• 2項目 $$\begin{aligned} &2\int_{0}^{1} x \left( x + \dfrac{1}{\sqrt{1+x^2}} \right) dx\\ &= 2\left[ \dfrac{x^3}{3} + \sqrt{1+x^2} \right]_0^1\\ &= -\dfrac{4}{3}+2\sqrt{2} \end{aligned}$$

以上をまとめると $$\begin{aligned} &\left( x^2 + \dfrac{x}{\sqrt{1+x^2}} \right) \left( 1 + \dfrac{x}{(1+x^2) \sqrt{1+x^2}} \right)\\ &= \left( -\dfrac{13}{12} + \dfrac{\sqrt{2}}{2} + \dfrac{\pi}{8} \right) + \left( -\dfrac{4}{3}+2\sqrt{2} \right)\\ &= -\dfrac{35}{12} + \dfrac{5}{2} \sqrt{2} + \dfrac{\pi}{8} \end{aligned}$$ となる。