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人気のある微分積分 >

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

解

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

解

\frac{3}{8} (3 x^{2} + 2 x)^{\frac{4}{3}} + C

解答ステップ

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

u置換積分法を適用する

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

べき乗則を適用する

= \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}}

代用を戻す

u = 3 x^{2} + 2 x

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

簡素化

\frac{1}{2} \cdot \frac{3}{4} (3 x^{2} + 2 x)^{\frac{4}{3}} : \frac{3}{8} (3 x^{2} + 2 x)^{\frac{4}{3}}

= \frac{3}{8} (3 x^{2} + 2 x)^{\frac{4}{3}}

定数を解答に追加する

= \frac{3}{8} (3 x^{2} + 2 x)^{\frac{4}{3}} + C

グラフ

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