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

The integral of xe^{-bx} is 1/(b^2)(-be^{-bx}x-e^{-bx})+C

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integral of xe^{-bx}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - b x

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

Simplify

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

Add a constant to the solution

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

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What is the integral of xe^{-bx} ?
The integral of xe^{-bx} is 1/(b^2)(-be^{-bx}x-e^{-bx})+C