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人気のある微分積分 >

integral of (9x(6x^2+5)^{1/2})/2

解

\int \frac{9 x ( 6 x ^{2} + 5 ) ^{\frac{1}{2}}}{2} d x

解

\frac{1}{4} (6 x^{2} + 5)^{\frac{3}{2}} + C

解答ステップ

\int \frac{9 x ( 6 x ^{2} + 5 ) ^{\frac{1}{2}}}{2} d x

定数を除く:

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

= \frac{9}{2} \cdot \int x (6 x^{2} + 5)^{\frac{1}{2}} d x

u置換積分法を適用する

= \frac{9}{2} \cdot \int \frac{( 6 u + 5 ) ^{\frac{1}{2}}}{2} d u

定数を除く:

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

= \frac{9}{2} \cdot \frac{1}{2} \cdot \int (6 u + 5)^{\frac{1}{2}} d u

u置換積分法を適用する

= \frac{9}{2} \cdot \frac{1}{2} \cdot \int \frac{v ^{\frac{1}{2}}}{6} d v

定数を除く:

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

= \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \int v^{\frac{1}{2}} d v

べき乗則を適用する

= \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{3} v^{\frac{3}{2}}

代用を戻す

= \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{3} (6 x^{2} + 5)^{\frac{3}{2}}

簡素化

\frac{9}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{3} (6 x^{2} + 5)^{\frac{3}{2}} : \frac{1}{4} (6 x^{2} + 5)^{\frac{3}{2}}

= \frac{1}{4} (6 x^{2} + 5)^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{4} (6 x^{2} + 5)^{\frac{3}{2}} + C

グラフ

人気の例

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\int \frac{1}{19 y} d y

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in v erse l a pl a ce \frac{4}{s ^{2} + 7 s + 12}

limit as x approaches 0 of e^x+1

x \to 0 lim (e^{x} + 1)

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derivative of (x^3+2x^{1/2}(x^{1/3}+5))

\frac{d}{d x} ((x^{3} + 2 x^{\frac{1}{2}}) (x^{\frac{1}{3}} + 5))