What is the integral of te^{-8t} ?

The integral of te^{-8t} is 1/64 (-8e^{-8t}t-e^{-8t})+C

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integral of te^{-8t}

\int t e^{- 8 t} d t

\frac{1}{64} (- 8 e^{- 8 t} t - e^{- 8 t}) + C

Solution steps

\int t e^{- 8 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{64} d u

= \int \frac{e ^{u} u}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{64} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{64} (e^{u} u - e^{u})

Substitute back

u = - 8 t

= \frac{1}{64} (e^{- 8 t} (- 8 t) - e^{- 8 t})

Simplify

= \frac{1}{64} (- 8 e^{- 8 t} t - e^{- 8 t})

Add a constant to the solution

= \frac{1}{64} (- 8 e^{- 8 t} t - e^{- 8 t}) + C

5 e x cos (x)

\frac{d y}{d x} = e^{x}, x (0) = 6

\frac{\partial}{\partial x} (x^{2} + y^{2} + (12 - x - y)^{2})

\frac{\partial}{\partial x} (x y e^{- y})

x \to l n (2) lim (- 4 e^{- 2 x})

What is the integral of te^{-8t} ?
The integral of te^{-8t} is 1/64 (-8e^{-8t}t-e^{-8t})+C