What is the integral of 9xe^{(-3x)} ?

The integral of 9xe^{(-3x)} is -3e^{-3x}x-e^{-3x}+C

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integral of 9xe^{(-3x)}

\int 9 x e^{(- 3 x)} d x

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 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

= 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 x

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

Simplify

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

= - 3 e^{- 3 x} x - e^{- 3 x}

Add a constant to the solution

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

\int ddd

in v erse l a pl a ce \frac{1}{4 s ^{2} + 1}

\int cos (\frac{1}{2}) x d x

\frac{d}{d x} (ln (2 x^{4} - x^{3} + 3 x^{2} - 3 x))

im pl i c i t \frac{d y}{d x}, x^{3} y^{5} + 3 x = 8 y^{3} + 1

What is the integral of 9xe^{(-3x)} ?
The integral of 9xe^{(-3x)} is -3e^{-3x}x-e^{-3x}+C