What is the integral of te^{-kt} ?

The integral of te^{-kt} is 1/(k^2)(-ke^{-kt}t-e^{-kt})+C

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integral of te^{-kt}

\int t e^{- k t} d t

\frac{1}{k ^{2}} (- k e^{- k t} t - e^{- k t}) + C

Solution steps

\int t e^{- k t} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - k t

= \frac{1}{k ^{2}} (e^{- k t} (- k t) - e^{- k t})

Simplify

= \frac{1}{k ^{2}} (- k e^{- k t} t - e^{- k t})

Add a constant to the solution

= \frac{1}{k ^{2}} (- k e^{- k t} t - e^{- k t}) + C

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What is the integral of te^{-kt} ?
The integral of te^{-kt} is 1/(k^2)(-ke^{-kt}t-e^{-kt})+C