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受欢迎的微积分 >

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

解答

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

解答

\frac{2}{15} (5 x^{\frac{2}{3}} - 8) (x^{\frac{2}{3}} + 2)^{\frac{5}{4}} + C

求解步骤

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

使用换元积分法

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

提出常数:

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

= 3 \cdot \int u^{3} (2 + u^{2})^{\frac{1}{4}} d u

使用换元积分法

= 3 \cdot \int \frac{( v - 2 ) v ^{\frac{1}{4}}}{2} d v

提出常数:

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

= 3 \cdot \frac{1}{2} \cdot \int (v - 2) v^{\frac{1}{4}} d v

乘开

(v - 2) v^{\frac{1}{4}} : v^{\frac{5}{4}} - 2 v^{\frac{1}{4}}

= 3 \cdot \frac{1}{2} \cdot \int v^{\frac{5}{4}} - 2 v^{\frac{1}{4}} d v

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{2} (\int v^{\frac{5}{4}} d v - \int 2 v^{\frac{1}{4}} d v)

\int v^{\frac{5}{4}} d v = \frac{4}{9} v^{\frac{9}{4}}

\int 2 v^{\frac{1}{4}} d v = \frac{8}{5} v^{\frac{5}{4}}

= 3 \cdot \frac{1}{2} (\frac{4}{9} v^{\frac{9}{4}} - \frac{8}{5} v^{\frac{5}{4}})

代回

= 3 \cdot \frac{1}{2} (\frac{4}{9} (2 + (x^{\frac{1}{3}})^{2})^{\frac{9}{4}} - \frac{8}{5} (2 + (x^{\frac{1}{3}})^{2})^{\frac{5}{4}})

化简

3 \cdot \frac{1}{2} (\frac{4}{9} (2 + (x^{\frac{1}{3}})^{2})^{\frac{9}{4}} - \frac{8}{5} (2 + (x^{\frac{1}{3}})^{2})^{\frac{5}{4}}) : \frac{2}{15} (5 x^{\frac{2}{3}} - 8) (x^{\frac{2}{3}} + 2)^{\frac{5}{4}}

= \frac{2}{15} (5 x^{\frac{2}{3}} - 8) (x^{\frac{2}{3}} + 2)^{\frac{5}{4}}

解答补常数

= \frac{2}{15} (5 x^{\frac{2}{3}} - 8) (x^{\frac{2}{3}} + 2)^{\frac{5}{4}} + C

作图

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