What is the integral of e^{-3t}t ?

The integral of e^{-3t}t is 1/9 (-3e^{-3t}t-e^{-3t})+C

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integral of e^{-3t}t

\int e^{- 3 t} t d t

\frac{1}{9} (- 3 e^{- 3 t} t - e^{- 3 t}) + C

Solution steps

\int e^{- 3 t} t d t

Apply u-substitution

= \int \frac{e ^{u} u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

= \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = - 3 t

= \frac{1}{9} (e^{- 3 t} (- 3 t) - e^{- 3 t})

Simplify

= \frac{1}{9} (- 3 e^{- 3 t} t - e^{- 3 t})

Add a constant to the solution

= \frac{1}{9} (- 3 e^{- 3 t} t - e^{- 3 t}) + C

What is the integral of e^{-3t}t ?
The integral of e^{-3t}t is 1/9 (-3e^{-3t}t-e^{-3t})+C