integral of ze^{2z}

解答

\int z e^{2 z} d z

\frac{1}{4} (e^{2 z} \cdot 2 z - e^{2 z}) + C

求解步骤

\int z e^{2 z} d z

使用换元积分法

= \int \frac{e ^{u} u}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int e^{u} u d u

使用分布积分法

= \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{4} (e^{u} u - e^{u})

u = 2 z

代回

= \frac{1}{4} (e^{2 z} \cdot 2 z - e^{2 z})

解答补常数

= \frac{1}{4} (e^{2 z} \cdot 2 z - e^{2 z}) + C

\int_{0}^{6} x^{3} x^{2} + 36 d x

x \to - 3 - lim (\frac{( x + 2 )}{x + 3})

x (e^{\frac{y}{x}} - 1) y^{^{'}} = e^{\frac{y}{x}} (y - x), y (1) = 0

\frac{d}{d x} ((x^{2} + y^{2}) ln (x + y))

\frac{d}{d x} (\frac{x + 3}{x - 5})