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問題：次の式を計算せよ

(1)の答え：\( 3^x\log3 + C \)
(2)の答え：\( \dfrac{2^x}{\log2}\left( x^2 - \dfrac{2x}{\log2} + \dfrac{2}{(\log2)^2} \right) + C \)
解説：
この積分は\( 2^x \)と\( x^2 \)の部分積分なので、瞬間部分積分を用いて解く。
(3)の答え：\( \sqrt{2} - 1 \)
解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{1} \dfrac{1}{x^2 + 1} dx & = & \displaystyle \int_{0}^{1} \dfrac{1}{\left( x + \dfrac{1}{2} \right)^2 - \dfrac{1}{4}} dx \\ & = & \displaystyle \int_{0}^{1} \dfrac{a}{x + \dfrac{3}{4}} + \dfrac{b}{x + \dfrac{1}{4}} dx \\ & = & -2\displaystyle \int_{0}^{1} \dfrac{1}{x + \dfrac{3}{4}} - \dfrac{1}{x + \dfrac{1}{4}} dx \\ & = & -2\left[ \log\left| x + \dfrac{3}{4} \right| - \log\left| x + \dfrac{1}{4} \right| \right]_{0}^{1}\\ & = & -2\left[ \log\left| \dfrac{x + \frac{3}{4}}{x + \frac{1}{4}} \right| \right]_{0}^{1} \\ & = & -2\left( \log\dfrac{7}{5} - \log3 \right) \\ & = & -2\left( \log7 - \log5 - \log3 \right) \end{eqnarray} \)

(4)の答え：\( \dfrac{1}{3}\log2 - \dfrac{\sqrt{3}}{9}\pi \)
解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{1} \dfrac{1}{x^3 + 1} dx & = & \displaystyle \int_{0}^{1} \dfrac{a}{x + 1} + \dfrac{bx + c}{x^2 - x + 1} dx \\ & = & \dfrac{1}{3}\displaystyle \int_{0}^{1} \dfrac{1}{x + 1} - \dfrac{x - 2}{x^2 - x + 1} dx \\ & = & \dfrac{1}{3}\left( \displaystyle \int_{0}^{1} \dfrac{1}{x + 1} dx - \int_{0}^{1} \dfrac{x - 2}{x^2 - x + 1} dx \right) \end{eqnarray} \)

ここで\( I_1 = \displaystyle \int_{0}^{1} \dfrac{1}{x + 1} dx \)、\( I_2 = \displaystyle \int_{0}^{1} \dfrac{x - 2}{x^2 - x + 1} dx \)とおくと

\( \hspace{35px} \begin{eqnarray} I_1 & = & \biggl[ \log|x + 1| \biggr]_{0}^{1} \\ & = & \log2 \\ I_2 & = & \dfrac{1}{2} \displaystyle \int_{0}^{1} \dfrac{2x - 1}{x^2 - x + 1} dx - \dfrac{3}{2} \displaystyle \int_{0}^{1} \dfrac{1}{x^2 - x + 1} dx \end{eqnarray} \)

ここで\( I_3 = \dfrac{1}{2} \displaystyle \int_{0}^{1} \dfrac{2x - 1}{x^2 - x + 1} dx \)、\( I_4 = \dfrac{3}{2} \displaystyle \int_{0}^{1} \dfrac{1}{x^2 - x + 1} dx \)とおくと

\( \hspace{35px} \begin{eqnarray} I_3 & = & \dfrac{1}{2}\biggl[ \log|x^2 - x + 1| \biggr]_{0}^{1} \\ & = & 0 \\ I_4 & = & \dfrac{3}{2} \displaystyle \int_{0}^{1} \dfrac{1}{\left( x - \dfrac{1}{2} \right)^2 + \dfrac{3}{4}} dx \\ & = & \dfrac{3}{2} \displaystyle \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \dfrac{1}{\dfrac{3}{4}\tan^2 \theta + \dfrac{3}{4}} \dfrac{\sqrt{3}}{2\cos^2 \theta} d\theta \\ & = & \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot \dfrac{\sqrt{3}}{2} \displaystyle \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} d\theta \\ & = & \sqrt{3}\bigl[ \theta \bigr]_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \\ & = & \dfrac{\sqrt{3}}{3}\pi \end{eqnarray} \)
\( \hspace{35px} \begin{eqnarray} \therefore I_2 & = & I_3 - I_4 \\ & = & -\dfrac{\sqrt{3}}{3}\pi \\ \end{eqnarray} \)
\( \hspace{35px} \begin{eqnarray} \therefore \displaystyle \int_{0}^{1} \dfrac{1}{x^3 + 1} dx & = & \dfrac{1}{3}\left( I_1 - I_2 \right) \\ & = & \dfrac{1}{3}\left( \log2 - \dfrac{\sqrt{3}}{3}\pi \right) \\ & = & \dfrac{1}{3}\log2 - \dfrac{\sqrt{3}}{9}\pi \end{eqnarray} \)
(5)の答え：\( \dfrac{\pi}{8} + \dfrac{5\sqrt{2}}{2} - \dfrac{35}{12} \)
解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{1} \left( x^2 + \dfrac{x}{\sqrt{1 + x^2}} \right) \biggl(1 + \dfrac{x}{(1 + x^2)\sqrt{1 + x^2}} \biggr) dx & = & \displaystyle \int_{0}^{1} x^2 + \dfrac{2x^3 + x}{(1 + x^2)\sqrt{1 + x^2}} + \dfrac{x^2}{(1 + x^2)^2} dx \\ \end{eqnarray} \)