What is the integral of 9te^{-3t} ?

The integral of 9te^{-3t} is -3e^{-3t}t-e^{-3t}+C

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integral of 9te^{-3t}

\int 9 t e^{- 3 t} d t

- 3 e^{- 3 t} t - e^{- 3 t} + C

Solution steps

\int 9 t e^{- 3 t} d t

Take the constant out:

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

= 9 \cdot \int t e^{- 3 t} d t

Apply u-substitution

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

= 9 \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - 3 t

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

Simplify

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

= - 3 e^{- 3 t} t - e^{- 3 t}

Add a constant to the solution

= - 3 e^{- 3 t} t - e^{- 3 t} + C

\int \frac{x}{( 3 x + 1 ) ^{\frac{1}{3}}} d x

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

\int \frac{5 x - 3}{x ^{2} + 5 x + 4} d x

What is the integral of 9te^{-3t} ?
The integral of 9te^{-3t} is -3e^{-3t}t-e^{-3t}+C