integral of x*e^{-ax}

解答

\int x \cdot e^{- a x} d x

\frac{1}{a ^{2}} (- a e^{- a x} x - e^{- a x}) + C

求解步骤

\int x e^{- a x} d x

使用换元积分法

= \int \frac{e ^{u} u}{a ^{2}} d u

提出常数:

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

= \frac{1}{a ^{2}} \cdot \int e^{u} u d u

使用分布积分法

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

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

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

u = - a x

代回

= \frac{1}{a ^{2}} (e^{- a x} (- a x) - e^{- a x})

化简

= \frac{1}{a ^{2}} (- a e^{- a x} x - e^{- a x})

解答补常数

= \frac{1}{a ^{2}} (- a e^{- a x} x - e^{- a x}) + C

\int \frac{x ^{2}}{( x ^{2} + 4 x + 5 ) ^{2}} d x

\frac{d}{d x} (\frac{1}{2 x + 5})

\int \frac{cos ^{2} ( x )}{1 - sin ( x )} d x

x \to 2 lim (9 x)

t an g e n t 5 x + 6, at x = 2