integral of e^{3x}x^3

解答

\int e^{3 x} x^{3} d x

\frac{1}{27} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2) + C

求解步骤

\int e^{3 x} x^{3} d x

使用换元积分法

= \int \frac{e ^{u} u ^{3}}{81} d u

提出常数:

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

= \frac{1}{81} \cdot \int e^{u} u^{3} d u

使用分布积分法

= \frac{1}{81} (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= \frac{1}{81} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

u = 3 x

代回

= \frac{1}{81} ((3 x)^{3} e^{3 x} - 3 ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x})))

化简

\frac{1}{81} ((3 x)^{3} e^{3 x} - 3 ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x}))) : \frac{1}{27} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

= \frac{1}{27} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

解答补常数

= \frac{1}{27} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2) + C

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