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人気のある微分積分 >

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

解

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

解

2 (\frac{2}{7} (1 + x)^{\frac{7}{2}} - \frac{4}{5} (1 + x)^{\frac{5}{2}} + \frac{2}{3} (1 + x)^{\frac{3}{2}}) + C

解答ステップ

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

u置換積分法を適用する

= \int 2 u^{2} 1 + u d u

定数を除く:

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

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

u置換積分法を適用する

= 2 \cdot \int (v - 1)^{2} v d v

拡張

(v - 1)^{2} v : v^{\frac{5}{2}} - 2 v^{\frac{3}{2}} + v

= 2 \cdot \int v^{\frac{5}{2}} - 2 v^{\frac{3}{2}} + 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

= 2 (\int v^{\frac{5}{2}} d v - \int 2 v^{\frac{3}{2}} d v + \int v d v)

\int v^{\frac{5}{2}} d v = \frac{2}{7} v^{\frac{7}{2}}

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

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

= 2 (\frac{2}{7} v^{\frac{7}{2}} - \frac{4}{5} v^{\frac{5}{2}} + \frac{2}{3} v^{\frac{3}{2}})

代用を戻す

= 2 (\frac{2}{7} (1 + x^{\frac{1}{2}})^{\frac{7}{2}} - \frac{4}{5} (1 + x^{\frac{1}{2}})^{\frac{5}{2}} + \frac{2}{3} (1 + x^{\frac{1}{2}})^{\frac{3}{2}})

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

= 2 (\frac{2}{7} (1 + x)^{\frac{7}{2}} - \frac{4}{5} (1 + x)^{\frac{5}{2}} + \frac{2}{3} (1 + x)^{\frac{3}{2}})

定数を解答に追加する

= 2 (\frac{2}{7} (1 + x)^{\frac{7}{2}} - \frac{4}{5} (1 + x)^{\frac{5}{2}} + \frac{2}{3} (1 + x)^{\frac{3}{2}}) + C

グラフ

人気の例

integral of 10^{2x}

\int 1 0^{2 x} d x

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x \to 0 lim (x + 3)

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\int 8 tan^{3} (2 x) sec^{5} (2 x) d x

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\frac{d}{d x} (e^{y} cos (x y))