What is the integral of 1/4 (e^{-3t}t) ?

The integral of 1/4 (e^{-3t}t) is 1/36 (-3e^{-3t}t-e^{-3t})+C

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integral of 1/4 (e^{-3t}t)

\int \frac{1}{4} (e^{- 3 t} t) d t

\frac{1}{36} (- 3 e^{- 3 t} t - e^{- 3 t}) + C

Solution steps

\int \frac{1}{4} e^{- 3 t} t d t

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{- 3 t} t d t

Apply u-substitution

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

= \frac{1}{4} \cdot \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}{4} \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{4} \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{4} \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = - 3 t

= \frac{1}{4} \cdot \frac{1}{9} (e^{- 3 t} (- 3 t) - e^{- 3 t})

Simplify

\frac{1}{4} \cdot \frac{1}{9} (e^{- 3 t} (- 3 t) - e^{- 3 t}) : \frac{1}{36} (- 3 e^{- 3 t} t - e^{- 3 t})

= \frac{1}{36} (- 3 e^{- 3 t} t - e^{- 3 t})

Add a constant to the solution

= \frac{1}{36} (- 3 e^{- 3 t} t - e^{- 3 t}) + C

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What is the integral of 1/4 (e^{-3t}t) ?
The integral of 1/4 (e^{-3t}t) is 1/36 (-3e^{-3t}t-e^{-3t})+C