tanの立方根の積分
tanの立方根の積分
\[ \int\sqrt[3]{\tan x}dx=\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C \]
\[ \int\sqrt[3]{\tan x}dx=\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C \]
\begin{align*}
\int\sqrt[3]{\tan x}dx & =3\int\frac{t^{3}}{t^{6}+1}dt\cmt{t=\sqrt[3]{\tan x}}\\
& =\frac{3}{2}\int\frac{u}{u^{3}+1}du\cmt{u=t^{2}}\\
& =\frac{1}{2}\int\left(\frac{u+1}{u^{2}-u+1}-\frac{1}{u+1}\right)du\\
& =\frac{1}{2}\int\left(\frac{\frac{1}{2}(u^{2}-u+1)'}{u^{2}-u+1}+\frac{\frac{3}{2}}{u^{2}-u+1}-\frac{1}{u+1}\right)du\\
& =\frac{1}{2}\int\left(\frac{\frac{1}{2}(u^{2}-u+1)'}{u^{2}-u+1}+\frac{\frac{3}{2}}{\left(u-\frac{1}{2}\right)^{2}+\frac{3}{4}}-\frac{1}{u+1}\right)du\\
& =\frac{1}{4}\log\left(u^{2}-u+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2u-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left|u+1\right|+C\\
& =\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C
\end{align*}
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\]