What is the integral of e^{-6x}9x ?

The integral of e^{-6x}9x is 1/4 (-6e^{-6x}x-e^{-6x})+C

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integral of e^{-6x}9x

\int e^{- 6 x} 9 x d x

\frac{1}{4} (- 6 e^{- 6 x} x - e^{- 6 x}) + C

Solution steps

\int e^{- 6 x} \cdot 9 x d x

Take the constant out:

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

= 9 \cdot \int e^{- 6 x} x d x

Apply u-substitution

: \int \frac{e ^{u} u}{36} d u

= 9 \cdot \int \frac{e ^{u} u}{36} d u

Take the constant out:

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

= 9 \cdot \frac{1}{36} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 9 \cdot \frac{1}{36} (e^{u} u - \int e^{u} d u)

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

= 9 \cdot \frac{1}{36} (e^{u} u - e^{u})

Substitute back

u = - 6 x

= 9 \cdot \frac{1}{36} (e^{- 6 x} (- 6 x) - e^{- 6 x})

Simplify

9 \cdot \frac{1}{36} (e^{- 6 x} (- 6 x) - e^{- 6 x}) : \frac{1}{4} (- 6 e^{- 6 x} x - e^{- 6 x})

= \frac{1}{4} (- 6 e^{- 6 x} x - e^{- 6 x})

Add a constant to the solution

= \frac{1}{4} (- 6 e^{- 6 x} x - e^{- 6 x}) + C

\frac{d}{d x} (5 x^{5} sin (2 x))

\frac{d}{d x} (e^{(0.5 x)})

f (x) = ∣ x - 5 ∣

x^{^{''}} - 2 x - 5 = 0

\frac{\partial}{\partial x} (e^{k x})

What is the integral of e^{-6x}9x ?
The integral of e^{-6x}9x is 1/4 (-6e^{-6x}x-e^{-6x})+C