What is the integral of te^{-5t} ?

The integral of te^{-5t} is 1/25 (-5e^{-5t}t-e^{-5t})+C

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integral of te^{-5t}

\int t e^{- 5 t} d t

\frac{1}{25} (- 5 e^{- 5 t} t - e^{- 5 t}) + C

Solution steps

\int t e^{- 5 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{25} d u

= \int \frac{e ^{u} u}{25} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{25} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{25} (e^{u} u - e^{u})

Substitute back

u = - 5 t

= \frac{1}{25} (e^{- 5 t} (- 5 t) - e^{- 5 t})

Simplify

= \frac{1}{25} (- 5 e^{- 5 t} t - e^{- 5 t})

Add a constant to the solution

= \frac{1}{25} (- 5 e^{- 5 t} t - e^{- 5 t}) + C

x \to 5 lim (x^{2} - 8 x + 15)

x \to 0 lim (\frac{2}{x})

f (x) = x^{2} (\frac{2}{x ^{2}} + \frac{5}{x ^{3}})

\int \frac{x ^{2} - 64}{x ^{4}} d x

\int_{0}^{4} \frac{5}{6 x ^{2} + 7 x + 1} d x

What is the integral of te^{-5t} ?
The integral of te^{-5t} is 1/25 (-5e^{-5t}t-e^{-5t})+C