2019年東大第1問の定積分の計算

2019年の東大数学(理系第1問)の定積分を少し変わった方法で計算してみたいと思います．


$${\displaystyle\int_0^1 \biggl(x^2+\frac{x}{\sqrt{x^2+1}}\biggr)\biggl(1+\frac{x}{\sqrt{x^2+1}(x^2+1)}\biggr) dx}$$

求める積分値を$${I}$$とおく．

$${\displaystyle\frac{x}{\sqrt{x^2+1}(x^2+1)}=\frac{d}{dx}\biggl(-\frac{1}{\sqrt{x^2+1}}\biggr)}$$ であることに注意すると

$${\displaystyle I=\int_0^1x\biggl(x+\frac{1}{\sqrt{x^2+1}}\biggr)\frac{d}{dx}\biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr) dx}$$

$${\displaystyle =\int_0^1x\biggl\{2x-\biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr)\biggr\} \frac{d}{dx}\biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr) dx}$$

$${\displaystyle =2\int_0^1x^2 \frac{d}{dx}\biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr) dx-\int_0^1x\biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr) \frac{d}{dx}\biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr) dx}$$

$${\displaystyle =2\biggl\{\biggl[x^2 \biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr)\biggr]_0^1-\int_0^12x \biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr)dx\biggr\}}$$

$${\displaystyle -\biggl\{\biggl[\frac{1}{2}x \biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr)^2 \biggr]_0^1-\int_0^1\frac{1}{2}\biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr)^2dx\biggr\}}$$

$${\displaystyle=2\biggl(1-\frac{1}{\sqrt{2}}\biggr)-4\int_0^1\biggl(x^2-\frac{x}{\sqrt{x^2+1}}\biggr) dx}$$

$${\displaystyle-\frac{1}{2}\biggl(1-\frac{1}{\sqrt{2}}\biggr)^2+\frac{1}{2}\int_0^1\biggl(x-\frac{1}{\sqrt{x^2+1}}\biggr)^2 dx}$$

$${\displaystyle=2-\sqrt{2}-4\biggl(\frac{1}{3}-\int_0^1\frac{x}{\sqrt{x^2+1}} dx\biggr) }$$

$${\displaystyle-\frac{1}{2}\biggl(\frac{3}{2}-\sqrt{2}\biggr)+\frac{1}{2}\int_0^1\biggl(x^2+\frac{1}{x^2+1}-\frac{2x}{\sqrt{x^2+1}}\biggr) dx}$$

$${\displaystyle=\frac{2}{3}-\sqrt{2}+4J-\frac{7}{12}+\frac{\sqrt{2}}{2}+\frac{\pi}{8}-J}$$

ただし $${J=\displaystyle\int_0^1\frac{x}{\sqrt{x^2+1}} dx}$$ とおき，$${\displaystyle\int_0^1\frac{1}{x^2+1} dx=\frac{\pi}{4}}$$ を用いた．

よって $${I=\displaystyle\frac{1}{12}-\frac{\sqrt{2}}{2}+\frac{\pi}{8}+3J}$$ である．

ここで $${J=\displaystyle\biggl[\sqrt{x^2+1}\biggr]_0^1=\sqrt{2}-1}$$ だから

$${\displaystyle I=\frac{5\sqrt{2}}{2}-\frac{35}{12}+\frac{\pi}{8}}$$ となる．