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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} + C

求解步骤

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

化简

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

= \int \frac{( 9 x ^{\frac{2}{3}} + 4 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

提出常数:

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

使用换元积分法

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

提出常数:

\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

使用幂法则

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

u = 9 x^{\frac{2}{3}} + 4

代回

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

化简

\frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} : \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

= \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

解答补常数

= \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} + C

作图

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