問題
\displaystyle\int_0^1 \dfrac{1}{x^3+1}\,dx
解答
\begin{align*}
\displaystyle\int_0^1 \dfrac{1}{x^3+1}\,dx
&=\displaystyle\int_0^1 \dfrac{1}{(x+1)(x^2-x+1)}\,dx \\[3mm]
&=\dfrac{1}{3}\displaystyle\int_0^1\biggl(\dfrac{1}{x+1}-\dfrac{x-2}{x^2-x+1}\biggr)\,dx \\[3mm]
&=\dfrac{1}{3}\displaystyle\int_0^1\biggl(\dfrac{1}{x+1}-\dfrac{1}{2}\cdot\dfrac{2x-1-3}{x^2-x+1}\biggr)\,dx \\[3mm]
&=\dfrac{1}{3}\displaystyle\int_0^1\biggl(\dfrac{1}{x+1}-\dfrac{1}{2}\cdot\dfrac{2x-1}{x^2-x+1}+\dfrac{3}{2}\cdot\dfrac{1}{x^2-x+1}\biggr)\,dx
\end{align*}
\begin{array}{l}
\begin{array}{rl}\bigcirc\kern-2.6mm{\raisebox{-.1mm}{1}}
\displaystyle\int_0^1 \dfrac{1}{x+1} \,dx
&=\biggl[\log |x+1| \biggr]_0^1 \\[3mm]
&=\log 2 \\[5mm]
\end{array} \\
\begin{array}{rl}\bigcirc\kern-2.6mm{\raisebox{-.1mm}{2}}
\displaystyle\int_0^1 \dfrac{2x-1}{x^2-x+1} \,dx
&=\biggl[\log |x^2-x+1|\biggr]_0^1 \\[5mm]
&=0 \\[5mm]
\end{array} \\
\begin{array}{rl}\bigcirc\kern-2.6mm{\raisebox{-.15mm}{3}}
\displaystyle\int_0^1 \dfrac{1}{x^2-x+1} \,dx
&=\displaystyle\int_0^1 \dfrac{1}{\biggl(x-\dfrac{1}{2}\biggr)^2+\dfrac{3}{4}}\,dx \\[10mm]
\end{array} \\
\begin{array}{rl}
x-\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}\tan\theta\,とおくと \\[3mm]
dx=\dfrac{\sqrt{3}}{2}\dfrac{1}{\cos^2\theta}\,d\theta \\[3mm]
\begin{array}{c|ccc}
t & 0 & \rightarrow & 1 \\\hline
\theta & -\frac{\pi}{6} & \rightarrow & \frac{\pi}{6}
\end{array} \\[5mm]
\end{array} \\
\begin{array}{rl}
\displaystyle\int_0^1 \dfrac{1}{x^2-x+1} \,dx
&=\displaystyle\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \dfrac{1}{\cfrac{3}{4}\,(\tan^2\theta+1)}\,\cdot\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{\cos^2\theta}\,d\theta \\[10mm]
&=\displaystyle\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\dfrac{2\sqrt{3}}{3}\,d\theta \\[5mm]
&=\biggl[\dfrac{2\sqrt{3}}{3}\theta\biggr]_{-\frac{\pi}{6}}^{\frac{\pi}{6}} =\dfrac{2\sqrt{3}}{3}\cdot\dfrac{\pi}{3}
=\dfrac{2\sqrt{3}}{9}\pi \\[10mm]
\end{array} \\[10mm]
\begin{array}{rl}
よって \\[3mm]
\displaystyle\int_0^1 \dfrac{1}{x^3+1}\,dx
&=\dfrac{1}{3}\log 2+\dfrac{1}{2}\cdot\dfrac{2\sqrt{3}}{9}\pi\\[3mm]
&=\dfrac{1}{3}\log 2+\dfrac{\sqrt{3}}{9}\pi
\end{array}
\end{array}
\begin{array}{rl}
\displaystyle\int_0^1 \dfrac{1}{x^3+1}\,dx
&=\dfrac{1}{3}\log 2+\dfrac{1}{2}\cdot\dfrac{2\sqrt{3}}{9}\pi\\[3mm]
&=\dfrac{1}{3}\log 2+\dfrac{\sqrt{3}}{9}\pi
\end{array}
