\[
a_1 = \int_{0}^{1}\frac{dx}{1+x} = \biggl[\log{\bigl|x+1\bigr|}\biggr]_{0}^{1} = \log{2}.
\]

変数変換 $x = \tan{\theta}$ を用いると，$x$ が $0$ から $1$ まで連続的に変化するとき，$\theta$ は $0$ から $\pi/4$ まで変化する．また
\[
\frac{dx}{d\theta} = 1 + \tan^2{\theta}
\]
であるから
\[
a_2 = \int_{0}^{1}\frac{dx}{1+x^2} = \int_{0}^{\pi/4} \frac{1}{1 + \tan^2{\theta}} \cdot (1 + \tan^2{\theta})\,d\theta = \Bigl[\theta\Bigr]_{0}^{\pi/4} = \frac{\pi}{4}.
\]

\[
\frac{1}{x^3+1} = \frac{1}{(x+1)(x^2-x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}
\]
を満たすような定数 $A,B,C$ を求める．両辺に $x^3+1 = (x+1)(x^2-x+1)$ を掛けて整理すると
\[
1 = A(x^2-x+1) + (Bx+C)(x+1) = (A+B)x^2 + (-A+B+C)x + A+C
\]
であるから，$A,B,C$ は次の連立方程式

$$
\begin{eqnarray*}
\begin{cases}
A+B = 0 & \\
-A+B+C = 0 & \\
A+C = 1
\end{cases}
\end{eqnarray*}
$$

を満たす．これを解いて
\[
A = \frac{1}{3}, \quad B = -\frac{1}{3}, \quad C = \frac{2}{3}
\]
を得るから，与えられた積分は

$$
\begin{eqnarray*}
a_3 & = & \int_{0}^{1}\frac{1}{1+x^3}\,dx \\
& = & \int_{0}^{1}\left(\frac{1}{3}\cdot\frac{1}{x+1} – \frac{1}{3}\cdot\frac{x-2}{x^2-x+1}\right)\,dx \\
& = & \frac{1}{3}\int_{0}^{1}\frac{dx}{x+1} – \frac{1}{3}\int_{0}^{1}\frac{x-2}{x^2-x+1}\,dx \\
& = & \frac{1}{3}\int_{0}^{1}\frac{dx}{x+1} – \frac{1}{6}\int_{0}^{1}\frac{2x-4}{x^2-x+1}\,dx \\
& = & \frac{1}{3}\int_{0}^{1}\frac{dx}{x+1} – \frac{1}{6}\int_{0}^{1}\frac{2x-1}{x^2-x+1}\,dx + \frac{1}{6}\int_{0}^{1}\frac{3}{x^2-x+1}\,dx \\
& = & \frac{1}{3}\int_{0}^{1}\frac{dx}{x+1} – \frac{1}{6}\int_{0}^{1}\frac{2x-1}{x^2-x+1}\,dx + \frac{1}{2}\int_{0}^{1}\frac{dx}{x^2-x+1}
\end{eqnarray*}
$$

と変形できる．ここで，それぞれの積分について考える．
\[
\int_{0}^{1}\frac{dx}{x+1} = \biggl[\log{|x+1|}\biggr]_{0}^{1} = \log{2},
\]
\[
\int_{0}^{1}\frac{2x-1}{x^2-x+1}\,dx = \biggl[\log{|x^2-x+1|}\biggr]_{0}^{1} = 0.
\]
最後の積分については，まず次の変形を行う．
\[
\int_{0}^{1}\frac{dx}{x^2-x+1} = \int_{0}^{1}\frac{dx}{(x-\frac{1}{2})^2+\frac{3}{4}}
\]
変数変換 $\displaystyle x – \frac{1}{2} = \frac{\sqrt{3}}{2}\tan{\theta}$ を考えると
\[
\int_{0}^{1}\frac{dx}{(x-\frac{1}{2})^2+\frac{3}{4}} = \int_{-\pi/6}^{\pi/6}\frac{2\sqrt{3}}{3}\,dx = \frac{2\sqrt{3}}{9}\pi
\]
を得る．したがって求める定積分は
\[
a_3 = \frac{1}{3}\cdot\log{2} -\frac{1}{6}\cdot 0 + \frac{1}{2}\cdot \frac{2\sqrt{3}}{9}\pi = \frac{\log{2}}{3} + \frac{\sqrt{3}}{9}\pi.
\]

$$
\begin{eqnarray*}
\frac{1}{x^4+1} & = & \frac{1}{x^4+1+2x^2-2x^2} \\
& = & \frac{1}{(x^2+1)^2-(\sqrt{2}x)^2} \\
& = & \frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} \\
& = & \frac{Ax+B}{x^2-\sqrt{2}x+1} + \frac{Cx+D}{x^2 + \sqrt{2}x+1}
\end{eqnarray*}
$$

を満たす定数 $A,B,C,D$ を求める．両辺に $x^4+1$ をかけて式を整理すると

$$
\begin{eqnarray*}
1 & = & (Ax+B)(x^2+\sqrt{2}x+1) + (Cx+D)(x^2-\sqrt{2}x+1) \\
& = & (A+C)x^3+(\sqrt{2}A+B-\sqrt{2}C+D)x^2 + (A + \sqrt{2}B + C -\sqrt{2}D)x + B+D
\end{eqnarray*}
$$

であるから，$A,B,C,D$ は次の連立方程式

$$
\begin{eqnarray*}
\begin{cases}
A+C = 0 & \\
\sqrt{2}A + B-\sqrt{2}C + D = 0 & \\
A+\sqrt{2}B + C-\sqrt{2}D = 0 & \\
B+D = 1
\end{cases}
\end{eqnarray*}
$$

を満たす．これを解いて
\[
A = -\frac{1}{2\sqrt{2}}, \quad B = \frac{1}{2}, \quad C = \frac{1}{2\sqrt{2}}, \quad D = \frac{1}{2}
\]
を得るから，与えられた積分は

$$
\begin{eqnarray*}
a_4 & = & \int_{0}^{1}\frac{1}{1+x^4}\,dx \\
& = & \int_{0}^{1}\frac{1}{4\sqrt{2}}\left(\frac{2x+2\sqrt{2}}{x^2+\sqrt{2}x+1} – \frac{2x-2\sqrt{2}}{x^2-\sqrt{2}x+1}\right)\,dx \\
& = & \int_{0}^{1}\frac{1}{4\sqrt{2}}\Biggl(\frac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1} + \frac{\sqrt{2}}{x^2+\sqrt{2}x+1}-\frac{2x-\sqrt{2}}{x^2-\sqrt{2}x+1}+ \frac{\sqrt{2}}{x^2-\sqrt{2}x+1}\Biggr)\,dx \\
& = & \frac{1}{4\sqrt{2}}\int_{0}^{1}\frac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1}\,dx + \frac{1}{4}\int_{0}^{1}\frac{dx}{x^2+\sqrt{2}x+1}-\frac{1}{4\sqrt{2}}\int_{0}^{1}\frac{2x-\sqrt{2}}{x^2-\sqrt{2}x+1}\,dx + \frac{1}{4}\int_{0}^{1}\frac{dx}{x^2-\sqrt{2}x+1}
\end{eqnarray*}
$$

と変形できる．それぞれの積分について考える．１つ目と３つ目は，
\[
\int_{0}^{1}\frac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1} = \biggl[\log{\bigl|x^2+\sqrt{2}x+1\bigr|}\biggr]_{0}^{1} = \log{(2+\sqrt{2})},
\]
\[
\int_{0}^{1}\frac{2x-\sqrt{2}}{x^2-\sqrt{2}x+1} = \biggl[\log{\bigl|x^2-\sqrt{2}x+1\bigr|}\biggr]_{0}^{1} = \log{(2-\sqrt{2})}
\]
である．２つ目と４つ目については，まず次の変形を行う．
\[
\int_{0}^{1}\frac{dx}{x^2+\sqrt{2}x+1} = \int_{0}^{1}\frac{dx}{(x+\frac{\sqrt{2}}{2})^2 + \frac{1}{2}},
\]
\[
\int_{0}^{1}\frac{dx}{x^2-\sqrt{2}x+1} = \int_{0}^{1}\frac{dx}{(x-\frac{\sqrt{2}}{2})^2 + \frac{1}{2}}.
\]
それぞれ変数変換 $\displaystyle x+\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\tan{\theta}$，$\displaystyle x-\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\tan{\theta}$ を用いれば
\[
\int_{0}^{1}\frac{dx}{(x+\frac{\sqrt{2}}{2})^2 + \frac{1}{2}} = \int_{\pi/4}^{3\pi/8}\sqrt{2}\,d\theta = \frac{\pi}{4\sqrt{2}},
\]
\[
\int_{0}^{1}\frac{dx}{(x-\frac{\sqrt{2}}{2})^2 + \frac{1}{2}} = \int_{-\pi/4}^{\pi/8}\sqrt{2}\,d\theta = \frac{3}{4\sqrt{2}}\pi
\]
を得る．ここで $x$ と $\theta$ の対応を考える際に
\[
\tan{\frac{\pi}{8}} = \sqrt{2}-1,　\quad \tan{\frac{3\pi}{8}} = \sqrt{2}+1
\]
を用いた．したがって求める定積分は

$$
\begin{eqnarray*}
a_4 & = & \frac{1}{4\sqrt{2}} \cdot \log{(2+\sqrt{2})} + \frac{1}{4} \cdot \frac{\pi}{4\sqrt{2}}-\frac{1}{4\sqrt{2}} \cdot \log{(2-\sqrt{2})} + \frac{1}{4} \cdot \frac{3}{4\sqrt{2}}\pi \\
& = & \frac{\log{(3+2\sqrt{2})} + \pi}{4\sqrt{2}}
\end{eqnarray*}
$$

である．ここで
\[
\log{(2+\sqrt{2})} -\log{(2-\sqrt{2})} = \log{\frac{2+\sqrt{2}}{2-\sqrt{2}}} = \log{(3+2\sqrt{2})}
\]
を用いた．

$a_3$ を求める際に導出した恒等式の $x$ を $x^2$ で置き換えて
\[
\frac{1}{1+x^6} = \frac{1}{3}\cdot\frac{1}{x^2+1}-\frac{1}{3}\cdot\frac{x^2-2}{x^4-x^2+1}
\]

を得る．第１項については
\[
\int_{0}^{1}\frac{dx}{x^2+1} = \frac{\pi}{4}
\]
である．第２項については，部分分数分解によって

$$
\begin{eqnarray*}
\frac{x^2-2}{x^4-x^2+1}
& = & \frac{\sqrt{3}}{4}\left(\frac{2x-\frac{4}{\sqrt{3}}}{x^2-\sqrt{3}x+1}-\frac{2x+\frac{4}{\sqrt{3}}}{x^2+\sqrt{3}x+1}\right) \\
& = & \frac{\sqrt{3}}{4}\cdot\frac{2x-\sqrt{3}}{x^2-\sqrt{3}x+1} – \frac{1}{4}\cdot\frac{1}{x^2-\sqrt{3}x+1}-\frac{\sqrt{3}}{4}\cdot\frac{2x+\sqrt{3}}{x^2+\sqrt{3}x+1}-\frac{1}{4}\cdot\frac{1}{x^2+\sqrt{3}x+1}
\end{eqnarray*}
$$

を得る．上式の第１，第３項の定積分は

\[
\int_{0}^{1}\frac{2x-\sqrt{3}}{x^2-\sqrt{3}x+1}\,dx = \biggl[\log{\bigl|x^2-\sqrt{3}x+1\bigr|}\biggr]_{0}^{1} = \log{(2-\sqrt{3})},
\]
\[
\int_{0}^{1}\frac{2x+\sqrt{3}}{x^2+\sqrt{3}x+1}\,dx = \biggl[\log{\bigl|x^2+\sqrt{3}x+1\bigr|}\biggr]_{0}^{1} = \log{(2+\sqrt{3})}
\]
である．第２，第４項については，まず次の変形を行う．
\[
\frac{1}{x^2-\sqrt{3}x+1} = \frac{1}{(x-\frac{\sqrt{3}}{2})^2+\frac{1}{4}}
\]
\[
\frac{1}{x^2+\sqrt{3}x+1} = \frac{1}{(x+\frac{\sqrt{3}}{2})^2+\frac{1}{4}}
\]

それぞれ変数変換 $\displaystyle x-\frac{\sqrt{3}}{2} = \frac{1}{2}\tan{\theta}$，$\displaystyle x+\frac{\sqrt{3}}{2} = \frac{1}{2}\tan{\theta}$ を用いれば
\[
\int_{0}^{1} \frac{1}{(x-\frac{\sqrt{3}}{2})^2+\frac{1}{4}}= \int_{-\pi/6}^{\pi/12}2\,d\theta = \frac{\pi}{2},
\]
\[
\int_{0}^{1}\frac{dx}{(x+\frac{\sqrt{3}}{2})^2 + \frac{1}{4}} = \int_{\pi/6}^{5\pi/12}2\,d\theta = \frac{\pi}{2}
\]
を得る．ただし $x$ と $\theta$ の対応を考える際に
\[
\tan{\frac{\pi}{12}} = 2-\sqrt{3}, \qquad \tan{\frac{5}{12}\pi} = 2 + \sqrt{3}
\]
を用いた．したがって求める定積分は

$$
\begin{eqnarray*}
a_6
& = & \frac{1}{3} \cdot \frac{\pi}{4}-\frac{1}{3}\left(\frac{\sqrt{3}}{4}\log{\frac{2-\sqrt{3}}{2+\sqrt{3}}}-\frac{\pi}{4}\right) \\
& = & \frac{\pi}{6} + \frac{\sqrt{3}}{12}\log{(7-4\sqrt{3})}
\end{eqnarray*}
$$

である．