integral of xe^{x+2}

解答

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

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

求解步骤

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

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

= \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

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

使用分布积分法

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

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

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

解答补常数

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

n = 1 \sum \infty \frac{( x - 1 ) ^{n}}{n}

d er i v a t i v e 6 t - 6 e^{t}

\int sin^{4} (y) cos^{7} (y) d y

im pl i c i t \frac{d ^{2} y}{d x ^{2}}, x^{2} + y^{2} = 36

\frac{d}{d x} (\frac{2 d}{x ^{3}} - a)