integral of ze^{4z^2}

解答

\int z e^{4 z^{2}} d z

\frac{1}{8} e^{4 z^{2}} + C

求解步骤

\int z e^{4 z^{2}} d z

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

= \frac{1}{8} e^{u}

u = 4 z^{2}

代回

= \frac{1}{8} e^{4 z^{2}}

解答补常数

= \frac{1}{8} e^{4 z^{2}} + C

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