What is the integral of 15e^{-31t}t ?

The integral of 15e^{-31t}t is 15/961 (-31e^{-31t}t-e^{-31t})+C

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integral of 15e^{-31t}t

\int 15 e^{- 31 t} t d t

\frac{15}{961} (- 31 e^{- 31 t} t - e^{- 31 t}) + C

Solution steps

\int 15 e^{- 31 t} t d t

Take the constant out:

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

= 15 \cdot \int e^{- 31 t} t d t

Apply u-substitution

: \int \frac{e ^{u} u}{961} d u

= 15 \cdot \int \frac{e ^{u} u}{961} d u

Take the constant out:

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

= 15 \cdot \frac{1}{961} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 15 \cdot \frac{1}{961} (e^{u} u - \int e^{u} d u)

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

= 15 \cdot \frac{1}{961} (e^{u} u - e^{u})

Substitute back

u = - 31 t

= 15 \cdot \frac{1}{961} (e^{- 31 t} (- 31 t) - e^{- 31 t})

Simplify

15 \cdot \frac{1}{961} (e^{- 31 t} (- 31 t) - e^{- 31 t}) : \frac{15}{961} (- 31 e^{- 31 t} t - e^{- 31 t})

= \frac{15}{961} (- 31 e^{- 31 t} t - e^{- 31 t})

Add a constant to the solution

= \frac{15}{961} (- 31 e^{- 31 t} t - e^{- 31 t}) + C

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What is the integral of 15e^{-31t}t ?
The integral of 15e^{-31t}t is 15/961 (-31e^{-31t}t-e^{-31t})+C