分母に2乗根と3乗根の積分
\[ \int\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx=2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\log\left(1+x^{\frac{1}{6}}\right)+C \]
\begin{align*} \int\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx & =\int\frac{1}{\left(x^{\frac{1}{6}}\right)^{3}+\left(x^{\frac{1}{6}}\right)^{2}}dx\\ & =6\int\frac{t^{5}}{t^{3}+t^{2}}dt\qquad,\qquad t=x^{\frac{1}{6}}\\ & =6\int\frac{t^{3}}{t+1}dt\\ & =6\int\left(t^{2}-t+1-\frac{1}{t+1}\right)dt\\ & =2t^{3}-3t^{2}+6t-6\log\left(1+t\right)+C\\ & =2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\log\left(1+x^{\frac{1}{6}}\right)+C \end{align*}
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sinの3乗をxの2乗で割った定積分
\[
\int_{0}^{\infty}\frac{\sin^{3}x}{x^{2}}dx=?
\]
分母に(1+x²)²を含む積分
\[
\int\frac{1}{\left(1+x^{2}\right)^{2}}dx=\frac{1}{2}\tan^{\bullet}x+\frac{x}{2\left(1+x^{2}\right)}+C
\]
tanの平方根の積分
\[
\int\sqrt{\tan x}dx=\frac{\sqrt{2}}{4}\log\left(\tan x-\sqrt{2\tan x}+1\right)-\frac{\sqrt{2}}{4}\log\left(\tan x+\sqrt{2\tan x}+1\right)+\frac{\sqrt{2}}{2}\tan^{\bullet}\left(\sqrt{2\tan x}-1\right)+\frac{\sqrt{2}}{2}\tan^{\bullet}\left(\sqrt{2\tan x}+1\right)+C
\]
tanの立方根の積分
\[
\int\sqrt[3]{\tan x}dx=\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C
\]