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受欢迎的微积分 >

integral of 6(e^{4-3x})^2

解答

\int 6 (e^{4 - 3 x})^{2} d x

解答

- e^{8 - 6 x} + C

求解步骤

\int 6 (e^{4 - 3 x})^{2} d x

提出常数:

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

= 6 \cdot \int (e^{4 - 3 x})^{2} d x

(e^{4 - 3 x})^{2} = (e^{4})^{2} (e^{- 3 x})^{2}

= 6 \cdot \int (e^{4})^{2} (e^{- 3 x})^{2} d x

提出常数:

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

= 6 (e^{4})^{2} \cdot \int (e^{- 3 x})^{2} d x

(e^{4})^{2} = e^{8}

= 6 e^{8} \cdot \int (e^{- 3 x})^{2} d x

化简

(e^{- 3 x})^{2} : e^{- 6 x}

= 6 e^{8} \cdot \int e^{- 6 x} d x

使用换元积分法

= 6 e^{8} \cdot \int - \frac{1}{6} e^{u} d u

提出常数:

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

= 6 e^{8} (- \frac{1}{6} \cdot \int e^{u} d u)

使用积分常见定则:

\int e^{u} d u = e^{u}

= 6 e^{8} (- \frac{1}{6} e^{u})

u = - 6 x

代回

= 6 e^{8} (- \frac{1}{6} e^{- 6 x})

化简

6 e^{8} (- \frac{1}{6} e^{- 6 x}) : - e^{8 - 6 x}

= - e^{8 - 6 x}

解答补常数

= - e^{8 - 6 x} + C

作图

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