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➔・高校生・受験生への挑戦状！積分最難問ランダム13題 | 受験の月,
・超難問積分13問【高校数学】｜kyoka
解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{2}^{4} \dfrac{x^2 + 1}{x^3 - x} dx & = & \displaystyle \int_{2}^{4} \dfrac{1 + \frac{1}{x^2}}{x - \frac{1}{x}} dx \\ & = & \left[ \log \biggl| x - \frac{1}{x} \biggr| \right]_{2}^{4} \\ & = & \log\frac{15}{4} - \log\frac{3}{2} \\ & = & \log\frac{5}{2} \end{eqnarray} \)解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int \dfrac{\cos x}{\sin^7 x - \sin x} dx & = & \displaystyle \int \dfrac{\frac{1}{\sin^7 x}}{1 - \frac{1}{\sin^6 x}} \cdot \cos x dx \\ & = & -\dfrac{1}{6}\log \biggl| 1 - \frac{1}{\sin^6 x} \biggr| + C \end{eqnarray} \)解説：
\( I = \displaystyle \int_{-1}^{1} \dfrac{2x^2(e^x\sin^2 x + \cos^2 x)}{(x^4 + x^2 + 1)(e^x + 1)} dx \) とする。
\( \displaystyle \int_{-1}^{1} \dfrac{2x^2(e^x\sin^2 x + \cos^2 x)}{(x^4 + x^2 + 1)(e^x + 1)} dx = \displaystyle \int_{-1}^{1} \dfrac{2x^2(e^{-x}\sin^2 x + \cos^2 x)}{(x^4 + x^2 + 1)(e^{-x} + 1)} dx \) より
\( \hspace{35px} \begin{eqnarray} 2I & = & \displaystyle \int_{-1}^{1} \dfrac{2x^2(e^x\sin^2 x + \cos^2 x)}{(x^4 + x^2 + 1)(e^x + 1)} dx + \displaystyle \int_{-1}^{1} \dfrac{2x^2(e^{-x}\sin^2 x + \cos^2 x)}{(x^4 + x^2 + 1)(e^{-x} + 1)} dx \\ & = & \displaystyle \int_{-1}^{1} \dfrac{2x^2(e^x\sin^2 x + \cos^2 x)}{(x^4 + x^2 + 1)(e^x + 1)} dx + \displaystyle \int_{-1}^{1} \dfrac{2x^2(\sin^2 x + e^x\cos^2 x)}{(x^4 + x^2 + 1)(e^x + 1)} dx \\ & = & \displaystyle \int_{-1}^{1} \dfrac{2x^2\{e^x(\sin^2 + \cos^2 x) + (\cos^2 + \sin^2 x)\}}{(x^4 + x^2 + 1)(e^x + 1)} dx \\ & = & \displaystyle \int_{-1}^{1} \dfrac{2x^2(e^x + 1)}{(x^4 + x^2 + 1)(e^x + 1)} dx \\ & = & \displaystyle \int_{-1}^{1} \dfrac{2x^2}{(x^4 + x^2 + 1)} dx \\ & = & \displaystyle \int_{-1}^{1} \biggl( \dfrac{x}{x^2 - x + 1} - \dfrac{x}{x^2 + x + 1} \biggr) dx \\ & = & \displaystyle \int_{-1}^{1} \biggl\{ \dfrac{1}{2}\biggl( \dfrac{2x - 1}{x^2 - x + 1} + \dfrac{1}{x^2 - x + 1} \biggr) - \dfrac{1}{2}\biggl( \dfrac{2x + 1}{x^2 + x + 1} - \dfrac{1}{x^2 + x + 1} \biggr) \biggr\} dx \\ & = & \dfrac{1}{2}\biggl( \left[ \log|x^2 - x + 1| - \log|x^2 + x + 1| \right]_{-1}^{1} + \displaystyle \int_{-1}^{1} \biggl( \dfrac{1}{x^2 - x + 1} + \dfrac{1}{x^2 + x + 1} \biggr) dx \biggr) \\ & = & \dfrac{1}{2}\biggl( -2\log 3 + \displaystyle \int_{-1}^{1} \dfrac{dx}{\Bigl( x - \dfrac{1}{2} \Bigr)^2 + \dfrac{3}{4}} + \displaystyle \int_{-1}^{1} \dfrac{dx}{\Bigl( x + \dfrac{1}{2} \Bigr)^2 + \dfrac{3}{4}} \biggr) \\ & = & \dfrac{1}{2}\biggl( -2\log 3 + \displaystyle \dfrac{4}{3}\dfrac{\sqrt{3}}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{6}} d\theta_1 + \displaystyle \dfrac{4}{3}\dfrac{\sqrt{3}}{2}\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} d\theta_2 \biggr) ( x - \dfrac{1}{2} = \dfrac{\sqrt{3}}{2}\tan \theta_1, x + \dfrac{1}{2} = \dfrac{\sqrt{3}}{2}\tan \theta_1 ) \\ & = & \dfrac{1}{2}\biggl( -2\log 3 + \displaystyle \dfrac{4}{3}\dfrac{\sqrt{3}}{2}\left[ \theta_1 \right]_{-\frac{\pi}{3}}^{\frac{\pi}{6}} + \displaystyle \dfrac{4}{3}\dfrac{\sqrt{3}}{2}\left[ \theta_2 \right]_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \biggr) \\ & = & \dfrac{1}{2}\biggl( -2\log 3 + \displaystyle 2 \cdot \dfrac{2}{\sqrt{3}}\Bigl( \dfrac{\pi}{6} + \dfrac{\pi}{3} \Bigr) \biggr) \\ & = & \dfrac{\pi}{\sqrt{3}} - \log 3 \\ \therefore I & = & \dfrac{\pi}{2\sqrt{3}} - \dfrac{1}{2}\log 3 \end{eqnarray} \)解説：
\( t = \sqrt{\tan x} \)とおくと、\( t^2 = \tan x \)
\( 2tdt = \dfrac{1}{\cos^2 x} dx \) より \( \dfrac{2t}{1 + \tan^2 x}dt = dx \Leftrightarrow \dfrac{2t}{1 + t^4}dt = dx \)
\( \hspace{35px} \begin{eqnarray}
\displaystyle \int_{0}^{\frac{\pi}{4}} \sqrt{\tan x} dx & = & \displaystyle \int_{0}^{\frac{\pi}{4}} t \cdot \dfrac{2t}{1 + t^4}dt \\
& = & \dfrac{\sqrt{2}}{4}\displaystyle \int_{0}^{1} \dfrac{2t - \sqrt{2}}{t^2 - \sqrt{2}t + 1} dt + \dfrac{1}{2}\displaystyle \int_{0}^{1} \dfrac{1}{t^2 - \sqrt{2}t + 1} dt - \dfrac{\sqrt{2}}{4}\displaystyle \int_{0}^{1} \dfrac{2t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} dt + \dfrac{1}{2}\displaystyle \int_{0}^{1} \dfrac{1}{t^2 + \sqrt{2}t + 1} dt \\
& = & \dfrac{\sqrt{2}}{4}\left[ \log\biggl| \dfrac{t^2 - \sqrt{2}t + 1}{t^2 + \sqrt{2}t + 1} \biggr| \right]_{0}^{1} + \dfrac{1}{2}\biggl( \displaystyle \int_{0}^{1} \dfrac{1}{t^2 - \sqrt{2}t + 1} dt + \displaystyle \int_{0}^{1} \dfrac{1}{t^2 + \sqrt{2}t + 1} dt \biggr) \\
& = & -\dfrac{\sqrt{2}}{2}\log(\sqrt{2} + 1) + \dfrac{1}{2}\biggl( \displaystyle \int_{0}^{1} \dfrac{1}{t^2 - \sqrt{2}t + 1} dt + \displaystyle \int_{0}^{1} \dfrac{1}{t^2 + \sqrt{2}t + 1} dt \biggr) \\
& = & -\dfrac{\sqrt{2}}{2}\log(\sqrt{2} + 1) + \dfrac{1}{2}\biggl( \displaystyle \int_{0}^{1} \dfrac{1}{(t - \frac{1}{\sqrt{2}})^2 + \frac{1}{2}} dt + \displaystyle \int_{0}^{1} \dfrac{1}{(t + \frac{1}{\sqrt{2}})^2 + \frac{1}{2}} dt \biggr) \\
& = & -\dfrac{\sqrt{2}}{2}\log(\sqrt{2} + 1) + \dfrac{1}{2}\biggl( \sqrt{2}\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{8}} d\theta + \sqrt{2}\displaystyle \int_{\frac{\pi}{4}}^{\frac{3}{8}\pi} d\phi \biggr) \Bigl( t - \frac{1}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\tan \theta, t + \frac{1}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\tan \phi \Bigr) \\
& = & -\dfrac{\sqrt{2}}{2}\log(\sqrt{2} + 1) + \dfrac{\sqrt{2}}{2}\biggl\{ \biggl( \dfrac{\pi}{8} - \biggl( -\dfrac{\pi}{4} \biggr) \biggr) + \biggl( \dfrac{3}{8}\pi - \dfrac{\pi}{4} \biggr) \biggr\} \\
& = & -\dfrac{\sqrt{2}}{2}\log(\sqrt{2} + 1) + \dfrac{\sqrt{2}}{2}\biggl( \dfrac{3}{8}\pi + \dfrac{\pi}{8} \biggr) \\
& = & -\dfrac{\sqrt{2}}{2}\log(\sqrt{2} + 1) + \dfrac{\sqrt{2}}{4}\pi \\
& = & \dfrac{\sqrt{2}}{4}\pi - \dfrac{\sqrt{2}}{2}\log(\sqrt{2} + 1)
\end{eqnarray} \)
解説：
\( \sqrt{-x^2 + x + 6} = \sqrt{(3 - x)(2 + x)} = (2 + x)\sqrt{\dfrac{3 - x}{2 + x}} \)
\( t = \sqrt{\dfrac{3 - x}{2 + x}} \)とおくと、\( (2 + x)t^2 = (3 - x) \Leftrightarrow (1 + t^2)x = 3 - 2t^2 \Leftrightarrow x = \dfrac{3 - 2t^2}{1 + t^2} \)
\( x + 2 = \dfrac{5}{1 + t^2} \) より \( dx = \dfrac{-10t}{(1 + t^2)^2} dt \)
\( \therefore \sqrt{-x^2 + x + 6} = \dfrac{5}{1 + t^2} \cdot t = \dfrac{5t}{1 + t^2}, 2x - 1 = \dfrac{5(1 - t^2)}{1 + t^2} \)