What is the integral of e^{-2t}t ?

The integral of e^{-2t}t is 1/4 (-2e^{-2t}t-e^{-2t})+C

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integral of e^{-2t}t

\int e^{- 2 t} t d t

\frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t}) + C

Solution steps

\int e^{- 2 t} t d t

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

= \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - 2 t

= \frac{1}{4} (e^{- 2 t} (- 2 t) - e^{- 2 t})

Simplify

= \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t})

Add a constant to the solution

= \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t}) + C

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{( e ^{- 3 x} )}{x}

\int \frac{( x ^{2} )}{x ^{2} + 64} d x

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

n = 0 \sum \infty \frac{4}{2 n + 1}

f (x) = 6 x e^{x}, (0, 0)

What is the integral of e^{-2t}t ?
The integral of e^{-2t}t is 1/4 (-2e^{-2t}t-e^{-2t})+C