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人気のある微分積分 >

integral of sqrt(1+(4/(3x^{1/3)))^2}

解

\int 1 + (\frac{4}{3 x ^{\frac{1}{3}}})^{2} d x

解

\frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} + C

解答ステップ

\int 1 + (\frac{4}{3 x ^{\frac{1}{3}}})^{2} d x

簡素化

1 + (\frac{4}{3 x ^{\frac{1}{3}}})^{2} : \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}}

= \int \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

定数を除く:

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

= \frac{1}{3} \cdot \int \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

u置換積分法を適用する

= \frac{1}{3} \cdot \int \frac{u ^{\frac{1}{2}}}{6} 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 \frac{1}{6} \cdot \int u^{\frac{1}{2}} d u

べき乗則を適用する

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

代用を戻す

u = 9 x^{\frac{2}{3}} + 16

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

簡素化

\frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} : \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

= \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} + C

グラフ

人気の例

derivative y=(f(x))/(x^2+2)

d er i v a t i v e y = \frac{f ( x )}{x ^{2} + 2}

(\partial)/(\partial L)(L^{1/4})

\frac{\partial}{\partial L} (L^{\frac{1}{4}})

integral of (2-8x)/(sqrt(81-4x^2))

\int \frac{2 - 8 x}{81 - 4 x ^{2}} d x

(dx)/(dt)=c_{1}x-c_{2}x^2

\frac{d x}{d t} = c_{1} x - c_{2} x^{2}

integral from-1 to 2 of 3/(x^2)

\int_{- 1}^{2} \frac{3}{x ^{2}} d x