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人気のある微分積分 >

integral of x^5sqrt(1+x^3)

解

\int x^{5} 1 + x^{3} d x

解

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

解答ステップ

\int x^{5} 1 + x^{3} d x

u置換積分法を適用する

= \int \frac{( u - 1 ) u}{3} d u

定数を除く:

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

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

拡張

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

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

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

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

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

代用を戻す

u = 1 + x^{3}

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

定数を解答に追加する

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

グラフ

人気の例

tangent x^2+4x-11=-2y^2-3y,(2,-1)

t an g e n t x^{2} + 4 x - 11 = - 2 y^{2} - 3 y, (2, - 1)

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

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

derivative of x^2+x+4

\frac{d}{d x} (x^{2} + x + 4)

(dh)/(dt)=4-16t^2,h(1)=0

\frac{d h}{d t} = 4 - 16 t^{2}, h (1) = 0

derivative of ln(x^2-16x)

\frac{d}{d x} (ln (x^{2} - 16 x))