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

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

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integral of te^{-3t}

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

= \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

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

x^{2} y^{^{''}} + 7 x y^{^{'}} + 5 y = 0

\frac{d}{d x} (\frac{2 + 3 x}{x ^{2}})

\int sin^{18} (x) cos^{3} (x) d x

g (x) = 4 x^{2} (x^{3} + 5 x)

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