What is the integral of te^{-4t} ?

The integral of te^{-4t} is 1/16 (-4e^{-4t}t-e^{-4t})+C

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integral of te^{-4t}

\int t e^{- 4 t} d t

\frac{1}{16} (- 4 e^{- 4 t} t - e^{- 4 t}) + C

Solution steps

\int t e^{- 4 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = - 4 t

= \frac{1}{16} (e^{- 4 t} (- 4 t) - e^{- 4 t})

Simplify

= \frac{1}{16} (- 4 e^{- 4 t} t - e^{- 4 t})

Add a constant to the solution

= \frac{1}{16} (- 4 e^{- 4 t} t - e^{- 4 t}) + C

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\frac{d}{d x} (\frac{12 x ^{6} - 3 x ^{2} + x ^{2}}{3 x ^{4}})

What is the integral of te^{-4t} ?
The integral of te^{-4t} is 1/16 (-4e^{-4t}t-e^{-4t})+C