What is the integral of xe^{-λx} ?

The integral of xe^{-λx} is 1/(λ^2)(-λe^{-λx}x-e^{-λx})+C

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integral of xe^{-λx}

\int x e^{- λ x} d x

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

Solution steps

\int x e^{- λ x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{λ ^{2}} d u

= \int \frac{e ^{u} u}{λ ^{2}} d u

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

= \frac{1}{λ ^{2}} (e^{u} u - e^{u})

Substitute back

u = - λ x

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

Simplify

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

Add a constant to the solution

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

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What is the integral of xe^{-λx} ?
The integral of xe^{-λx} is 1/(λ^2)(-λe^{-λx}x-e^{-λx})+C