What is the integral of te^{-6t} ?

The integral of te^{-6t} is 1/36 (-6e^{-6t}t-e^{-6t})+C

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integral of te^{-6t}

\int t e^{- 6 t} d t

\frac{1}{36} (- 6 e^{- 6 t} t - e^{- 6 t}) + C

Solution steps

\int t e^{- 6 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{36} d u

= \int \frac{e ^{u} u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{36} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{36} (e^{u} u - e^{u})

Substitute back

u = - 6 t

= \frac{1}{36} (e^{- 6 t} (- 6 t) - e^{- 6 t})

Simplify

= \frac{1}{36} (- 6 e^{- 6 t} t - e^{- 6 t})

Add a constant to the solution

= \frac{1}{36} (- 6 e^{- 6 t} t - e^{- 6 t}) + C

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 6}{x})

n = 1 \sum \infty 4 (0.8)^{n - 1}

d x + e^{3 x} d y = 0

\frac{d ^{2}}{d x ^{2}} (- \frac{5}{x})

\frac{d}{d x} (\frac{- 10}{x ^{3}})

What is the integral of te^{-6t} ?
The integral of te^{-6t} is 1/36 (-6e^{-6t}t-e^{-6t})+C