What is the integral of te^{-9t} ?

The integral of te^{-9t} is 1/81 (-9e^{-9t}t-e^{-9t})+C

Popular >

integral of te^{-9t}

\int t e^{- 9 t} d t

\frac{1}{81} (- 9 e^{- 9 t} t - e^{- 9 t}) + C

Solution steps

\int t e^{- 9 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{81} d u

= \int \frac{e ^{u} u}{81} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{81} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{81} (e^{u} u - e^{u})

Substitute back

u = - 9 t

= \frac{1}{81} (e^{- 9 t} (- 9 t) - e^{- 9 t})

Simplify

= \frac{1}{81} (- 9 e^{- 9 t} t - e^{- 9 t})

Add a constant to the solution

= \frac{1}{81} (- 9 e^{- 9 t} t - e^{- 9 t}) + C

\int \frac{e ^{2 x}}{1 + e ^{2 x}} d x

\int_{0}^{π} 5 sin^{2} (x) d x

\int \frac{ln ( x ^{\frac{3}{2}} )}{x} d x

\int_{0}^{t} 3 e^{s} sin (t - s) d s

\int \frac{1}{2 x + 3} d x

What is the integral of te^{-9t} ?
The integral of te^{-9t} is 1/81 (-9e^{-9t}t-e^{-9t})+C