integral of 4xe^{x+2}

解答

\int 4 x e^{x + 2} d x

4 e^{2} (e^{x} x - e^{x}) + C

求解步骤

\int 4 x e^{x + 2} d x

提出常数:

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

= 4 \cdot \int x e^{x + 2} d x

x e^{x + 2} = e^{2} x e^{x}

= 4 \cdot \int e^{2} x e^{x} d x

提出常数:

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

= 4 e^{2} \cdot \int x e^{x} d x

使用分布积分法

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

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

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

解答补常数

= 4 e^{2} (e^{x} x - e^{x}) + C

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