What is the integral of xe^{-sx} ?

The integral of xe^{-sx} is 1/(s^2)(-se^{-sx}x-e^{-sx})+C

integral of xe^{-sx}

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

Substitute back

u = - s x

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (8 x^{5} y^{3} - 9 x^{4} y^{6})

\frac{d}{d x} (\frac{4 x ^{3} + 1}{x ^{2} - x + 3})

\frac{d y}{d x} - \frac{y}{x} = \frac{x ^{2} + y ^{2}}{x}

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

\frac{d}{d x} (x + 7)

What is the integral of xe^{-sx} ?
The integral of xe^{-sx} is 1/(s^2)(-se^{-sx}x-e^{-sx})+C