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人気のある微分積分 >

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

解

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

解

\frac{3}{40} (5 x^{2} - 4 x + 4) (3 x + 2)^{\frac{2}{3}} + C

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

= 3 \cdot \int \frac{( u - 2 ) ^{2}}{27 3 u} d u

定数を除く:

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

= 3 \cdot \frac{1}{27} \cdot \int \frac{( u - 2 ) ^{2}}{3 u} d u

拡張

\frac{( u - 2 ) ^{2}}{3 u} : u^{\frac{5}{3}} - 4 u^{\frac{2}{3}} + \frac{4}{3 u}

= 3 \cdot \frac{1}{27} \cdot \int u^{\frac{5}{3}} - 4 u^{\frac{2}{3}} + \frac{4}{3 u} 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

= 3 \cdot \frac{1}{27} (\int u^{\frac{5}{3}} d u - \int 4 u^{\frac{2}{3}} d u + \int \frac{4}{3 u} d u)

\int u^{\frac{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

\int 4 u^{\frac{2}{3}} d u = \frac{12}{5} u^{\frac{5}{3}}

\int \frac{4}{3 u} d u = 6 u^{\frac{2}{3}}

= 3 \cdot \frac{1}{27} (\frac{3}{8} u^{\frac{8}{3}} - \frac{12}{5} u^{\frac{5}{3}} + 6 u^{\frac{2}{3}})

代用を戻す

u = 2 + 3 x

= 3 \cdot \frac{1}{27} (\frac{3}{8} (2 + 3 x)^{\frac{8}{3}} - \frac{12}{5} (2 + 3 x)^{\frac{5}{3}} + 6 (2 + 3 x)^{\frac{2}{3}})

簡素化

3 \cdot \frac{1}{27} (\frac{3}{8} (2 + 3 x)^{\frac{8}{3}} - \frac{12}{5} (2 + 3 x)^{\frac{5}{3}} + 6 (2 + 3 x)^{\frac{2}{3}}) : \frac{3}{40} (5 x^{2} - 4 x + 4) (3 x + 2)^{\frac{2}{3}}

= \frac{3}{40} (5 x^{2} - 4 x + 4) (3 x + 2)^{\frac{2}{3}}

定数を解答に追加する

= \frac{3}{40} (5 x^{2} - 4 x + 4) (3 x + 2)^{\frac{2}{3}} + C

グラフ

人気の例

integral of 1/((9x^2+16)^{3/2)}

\int \frac{1}{( 9 x ^{2} + 16 ) ^{\frac{3}{2}}} d x

integral of (x^2+3x-5)/(x-1)

\int \frac{x ^{2} + 3 x - 5}{x - 1} d x

integral of x/(5+2x)

\int \frac{x}{5 + 2 x} d x

integral of 10x^2cos(9x)

\int 10 x^{2} cos (9 x) d x

integral of x^4 1/(sqrt(1-x^2))

\int x^{4} \frac{1}{1 - x ^{2}} d x