What is the integral of e^{-sx}*x ?

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

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integral of e^{-sx}*x

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

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

Solution steps

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

Apply u-substitution

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

= \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

: e^{u} u - \int e^{u} d u

= \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

f (x) = \frac{x ^{2} - 9}{x ^{2} + 3 x}

y = x, y = 2 x 100 - x^{2}

f (x) = 186.7

y^{^{''}} - y^{^{'}} + y = 3 sin (t) - 15 e^{2 t}

\frac{d}{d x} ((9 x - 1)^{- 2})

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