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受欢迎的微积分 >

integral of x\sqrt[4]{x+20}

解答

\int x 4 x + 20 d x

解答

4 (\frac{1}{9} (x + 20)^{\frac{9}{4}} - 4 (x + 20)^{\frac{5}{4}}) + C

求解步骤

\int x 4 x + 20 d x

使用换元积分法

= \int 4 u^{4} (u^{4} - 20) d u

提出常数:

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

= 4 \cdot \int u^{4} (u^{4} - 20) d u

乘开

u^{4} (u^{4} - 20) : u^{8} - 20 u^{4}

= 4 \cdot \int u^{8} - 20 u^{4} d u

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

= 4 (\int u^{8} d u - \int 20 u^{4} d u)

\int u^{8} d u = \frac{u ^{9}}{9}

\int 20 u^{4} d u = 4 u^{5}

= 4 (\frac{u ^{9}}{9} - 4 u^{5})

u = 4 x + 20

代回

= 4 (\frac{( 4 x + 20 ) ^{9}}{9} - 4 (4 x + 20)^{5})

化简

4 (\frac{( 4 x + 20 ) ^{9}}{9} - 4 (4 x + 20)^{5}) : 4 (\frac{1}{9} (x + 20)^{\frac{9}{4}} - 4 (x + 20)^{\frac{5}{4}})

= 4 (\frac{1}{9} (x + 20)^{\frac{9}{4}} - 4 (x + 20)^{\frac{5}{4}})

解答补常数

= 4 (\frac{1}{9} (x + 20)^{\frac{9}{4}} - 4 (x + 20)^{\frac{5}{4}}) + C

作图

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