与えられた定積分 $\int_{0}^{t} \left( \frac{\frac{1}{2\sqrt{2}}x}{x^2-\sqrt{2}x+1} - \frac{\frac{1}{2\sqrt{2}}x}{x^2+\sqrt{2}x+1} \right) dx$ を計算する。

1. 問題の内容

与えられた定積分

$\int_{0}^{t} \left( \frac{\frac{1}{2\sqrt{2}}x}{x^2-\sqrt{2}x+1} - \frac{\frac{1}{2\sqrt{2}}x}{x^2+\sqrt{2}x+1} \right) dx$

を計算する。

2. 解き方の手順

積分を分解して、各項を計算する。

$\int_{0}^{t} \left( \frac{\frac{1}{2\sqrt{2}}x}{x^2-\sqrt{2}x+1} - \frac{\frac{1}{2\sqrt{2}}x}{x^2+\sqrt{2}x+1} \right) dx = \frac{1}{2\sqrt{2}} \int_{0}^{t} \left( \frac{x}{x^2-\sqrt{2}x+1} - \frac{x}{x^2+\sqrt{2}x+1} \right) dx$

それぞれの分数の分子を、分母の微分形にする。

$x^2-\sqrt{2}x+1$ の微分は $2x-\sqrt{2}$ なので、$x=\frac{1}{2}(2x-\sqrt{2}) + \frac{\sqrt{2}}{2}$ と変形する。

$x^2+\sqrt{2}x+1$ の微分は $2x+\sqrt{2}$ なので、$x=\frac{1}{2}(2x+\sqrt{2}) - \frac{\sqrt{2}}{2}$ と変形する。

$\frac{1}{2\sqrt{2}} \int_{0}^{t} \left( \frac{\frac{1}{2}(2x-\sqrt{2}) + \frac{\sqrt{2}}{2}}{x^2-\sqrt{2}x+1} - \frac{\frac{1}{2}(2x+\sqrt{2}) - \frac{\sqrt{2}}{2}}{x^2+\sqrt{2}x+1} \right) dx$

$= \frac{1}{2\sqrt{2}} \int_{0}^{t} \left( \frac{1}{2} \frac{2x-\sqrt{2}}{x^2-\sqrt{2}x+1} + \frac{\sqrt{2}}{2} \frac{1}{x^2-\sqrt{2}x+1} - \frac{1}{2} \frac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1} + \frac{\sqrt{2}}{2} \frac{1}{x^2+\sqrt{2}x+1} \right) dx$

$= \frac{1}{2\sqrt{2}} \left[ \frac{1}{2} \ln(x^2-\sqrt{2}x+1) + \frac{\sqrt{2}}{2} \int_{0}^{t} \frac{1}{x^2-\sqrt{2}x+1} dx - \frac{1}{2} \ln(x^2+\sqrt{2}x+1) + \frac{\sqrt{2}}{2} \int_{0}^{t} \frac{1}{x^2+\sqrt{2}x+1} dx \right]_0^t$

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

$x^2 + \sqrt{2}x + 1 = (x+\frac{\sqrt{2}}{2})^2 + \frac{1}{2}$

$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan(\frac{x}{a})$ より

$\int_{0}^{t} \frac{1}{x^2-\sqrt{2}x+1} dx = \int_{0}^{t} \frac{1}{(x-\frac{\sqrt{2}}{2})^2 + \frac{1}{2}} dx = \sqrt{2} \arctan(\sqrt{2}x-1) \vert_0^t = \sqrt{2} (\arctan(\sqrt{2}t-1) - \arctan(-1)) = \sqrt{2} (\arctan(\sqrt{2}t-1) + \frac{\pi}{4})$

$\int_{0}^{t} \frac{1}{x^2+\sqrt{2}x+1} dx = \int_{0}^{t} \frac{1}{(x+\frac{\sqrt{2}}{2})^2 + \frac{1}{2}} dx = \sqrt{2} \arctan(\sqrt{2}x+1) \vert_0^t = \sqrt{2} (\arctan(\sqrt{2}t+1) - \arctan(1)) = \sqrt{2} (\arctan(\sqrt{2}t+1) - \frac{\pi}{4})$

$= \frac{1}{4\sqrt{2}} \left[ \ln \frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1} + \sqrt{2} ( \sqrt{2} \arctan(\sqrt{2}x-1) + \sqrt{2} \arctan(\sqrt{2}x+1) \right]_0^t$

$= \frac{1}{4\sqrt{2}} \left( \ln \frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1} + 2 \arctan(\sqrt{2}t-1) + 2 \arctan(\sqrt{2}t+1) - 0 - 2(\arctan(-1)+\arctan(1)) \right)$

$= \frac{1}{4\sqrt{2}} \left( \ln \frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1} + 2 \arctan(\sqrt{2}t-1) + 2 \arctan(\sqrt{2}t+1) \right)$

$\arctan(a) + \arctan(b) = \arctan(\frac{a+b}{1-ab})$

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

$= \frac{1}{4\sqrt{2}} \left( \ln \frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1} + 2 \arctan(\frac{\sqrt{2}t}{1-t^2}) \right)$

$=\frac{1}{4\sqrt{2}}\ln\left(\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right)+\frac{1}{2\sqrt{2}}\arctan\left(\frac{\sqrt{2}t}{1-t^2}\right)$

3. 最終的な答え

$\frac{1}{4\sqrt{2}}\ln\left(\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right)+\frac{1}{2\sqrt{2}}\arctan\left(\frac{\sqrt{2}t}{1-t^2}\right)$