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人気のある微分積分 >

integral of 4/(x^{1/3)+x^{1/2}}

解

\int \frac{4}{x ^{\frac{1}{3}} + x ^{\frac{1}{2}}} d x

解

24 (\frac{1}{3} (1 + x^{\frac{1}{6}})^{3} - \frac{3}{2} (1 + x^{\frac{1}{6}})^{2} + 3 (1 + x^{\frac{1}{6}}) - ln 1 + x^{\frac{1}{6}}) + C

解答ステップ

\int \frac{4}{x ^{\frac{1}{3}} + x ^{\frac{1}{2}}} d x

定数を除く:

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

= 4 \cdot \int \frac{1}{x ^{\frac{1}{3}} + x ^{\frac{1}{2}}} d x

u置換積分法を適用する

= 4 \cdot \int \frac{6 u ^{3}}{1 + u} d u

定数を除く:

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

= 4 \cdot 6 \cdot \int \frac{u ^{3}}{1 + u} d u

u置換積分法を適用する

= 4 \cdot 6 \cdot \int \frac{( v - 1 ) ^{3}}{v} d v

拡張

\frac{( v - 1 ) ^{3}}{v} : v^{2} - 3 v + 3 - \frac{1}{v}

= 4 \cdot 6 \cdot \int v^{2} - 3 v + 3 - \frac{1}{v} d v

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

= 4 \cdot 6 (\int v^{2} d v - \int 3 v d v + \int 3 d v - \int \frac{1}{v} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int 3 v d v = \frac{3 v ^{2}}{2}

\int 3 d v = 3 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 4 \cdot 6 (\frac{v ^{3}}{3} - \frac{3 v ^{2}}{2} + 3 v - ln ∣ v ∣)

代用を戻す

= 4 \cdot 6 \frac{( 1 + x ^{\frac{1}{6}} ) ^{3}}{3} - \frac{3 ( 1 + x ^{\frac{1}{6}} ) ^{2}}{2} + 3 (1 + x^{\frac{1}{6}}) - ln 1 + x^{\frac{1}{6}}

簡素化

4 \cdot 6 \frac{( 1 + x ^{\frac{1}{6}} ) ^{3}}{3} - \frac{3 ( 1 + x ^{\frac{1}{6}} ) ^{2}}{2} + 3 (1 + x^{\frac{1}{6}}) - ln 1 + x^{\frac{1}{6}} : 24 (\frac{1}{3} (1 + x^{\frac{1}{6}})^{3} - \frac{3}{2} (1 + x^{\frac{1}{6}})^{2} + 3 (1 + x^{\frac{1}{6}}) - ln 1 + x^{\frac{1}{6}})

= 24 (\frac{1}{3} (1 + x^{\frac{1}{6}})^{3} - \frac{3}{2} (1 + x^{\frac{1}{6}})^{2} + 3 (1 + x^{\frac{1}{6}}) - ln 1 + x^{\frac{1}{6}})

定数を解答に追加する

= 24 (\frac{1}{3} (1 + x^{\frac{1}{6}})^{3} - \frac{3}{2} (1 + x^{\frac{1}{6}})^{2} + 3 (1 + x^{\frac{1}{6}}) - ln 1 + x^{\frac{1}{6}}) + C

グラフ

人気の例

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