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

The integral of te^{-at} is 1/(a^2)(-ae^{-at}t-e^{-at})+C

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integral of te^{-at}

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

Substitute back

u = - a t

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

Simplify

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

Add a constant to the solution

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

What is the integral of te^{-at} ?
The integral of te^{-at} is 1/(a^2)(-ae^{-at}t-e^{-at})+C