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人気のある微分積分 >

integral of x^2sqrt(2x+1)

解

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

解

\frac{1}{105} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}} + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{u ( u - 1 ) ^{2}}{8} d u

定数を除く:

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

= \frac{1}{8} \cdot \int u (u - 1)^{2} d u

拡張

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

= \frac{1}{8} \cdot \int u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + 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

= \frac{1}{8} (\int u^{\frac{5}{2}} d u - \int 2 u^{\frac{3}{2}} d u + \int u d u)

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

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

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

= \frac{1}{8} (\frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

代用を戻す

u = 2 x + 1

= \frac{1}{8} (\frac{2}{7} (2 x + 1)^{\frac{7}{2}} - \frac{4}{5} (2 x + 1)^{\frac{5}{2}} + \frac{2}{3} (2 x + 1)^{\frac{3}{2}})

簡素化

\frac{1}{8} (\frac{2}{7} (2 x + 1)^{\frac{7}{2}} - \frac{4}{5} (2 x + 1)^{\frac{5}{2}} + \frac{2}{3} (2 x + 1)^{\frac{3}{2}}) : \frac{1}{105} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}}

= \frac{1}{105} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{105} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}} + C

グラフ

人気の例

limit as x approaches 0-of x+1

x \to 0 - lim (x + 1)

面積 y=1,x=y^{5/2}

a re a y = 1, x = y^{\frac{5}{2}}

inverselaplace (3s-5)/(s-1)

in v erse l a pl a ce \frac{3 s - 5}{s - 1}

derivative of ln(\sqrt[3]{x+1})

\frac{d}{d x} (ln (3 x + 1))

(x^{3} (8 - 3 x))^{^{'}}