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人気のある微分積分 >

integral of \sqrt[3]{x^2}(2x^2+1/(x^3))

解

\int 3 x^{2} (2 x^{2} + \frac{1}{x ^{3}}) d x

解

\frac{24 x ^{5} - 33}{44 x ^{\frac{4}{3}}} + C

解答ステップ

\int 3 x^{2} (2 x^{2} + \frac{1}{x ^{3}}) d x

u置換積分法を適用する

= \int \frac{2 u ^{\frac{5}{2}} + 1}{2 u ^{\frac{5}{3}}} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{2 u ^{\frac{5}{2}} + 1}{u ^{\frac{5}{3}}} d u

拡張

\frac{2 u ^{\frac{5}{2}} + 1}{u ^{\frac{5}{3}}} : 2 u^{\frac{5}{6}} + \frac{1}{u ^{\frac{5}{3}}}

= \frac{1}{2} \cdot \int 2 u^{\frac{5}{6}} + \frac{1}{u ^{\frac{5}{3}}} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 2 u^{\frac{5}{6}} d u + \int \frac{1}{u ^{\frac{5}{3}}} d u)

\int 2 u^{\frac{5}{6}} d u = \frac{12}{11} u^{\frac{11}{6}}

\int \frac{1}{u ^{\frac{5}{3}}} d u = - \frac{3}{2 u ^{\frac{2}{3}}}

= \frac{1}{2} (\frac{12}{11} u^{\frac{11}{6}} - \frac{3}{2 u ^{\frac{2}{3}}})

代用を戻す

u = x^{2}

= \frac{1}{2} (\frac{12}{11} (x^{2})^{\frac{11}{6}} - \frac{3}{2 ( x ^{2} ) ^{\frac{2}{3}}})

簡素化

\frac{1}{2} (\frac{12}{11} (x^{2})^{\frac{11}{6}} - \frac{3}{2 ( x ^{2} ) ^{\frac{2}{3}}}) : \frac{24 x ^{5} - 33}{44 x ^{\frac{4}{3}}}

= \frac{24 x ^{5} - 33}{44 x ^{\frac{4}{3}}}

定数を解答に追加する

= \frac{24 x ^{5} - 33}{44 x ^{\frac{4}{3}}} + C

グラフ

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