What is the integral of 3xe^{(-2x)} ?

The integral of 3xe^{(-2x)} is 3/4 (-2e^{-2x}x-e^{-2x})+C

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integral of 3xe^{(-2x)}

\int 3 x e^{(- 2 x)} d x

\frac{3}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + C

Solution steps

\int 3 x e^{- 2 x} d x

Take the constant out:

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

= 3 \cdot \int x e^{- 2 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

= 3 \cdot \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= 3 \cdot \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = - 2 x

= 3 \cdot \frac{1}{4} (e^{- 2 x} (- 2 x) - e^{- 2 x})

Simplify

3 \cdot \frac{1}{4} (e^{- 2 x} (- 2 x) - e^{- 2 x}) : \frac{3}{4} (- 2 e^{- 2 x} x - e^{- 2 x})

= \frac{3}{4} (- 2 e^{- 2 x} x - e^{- 2 x})

Add a constant to the solution

= \frac{3}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + C

\frac{\partial}{\partial y} (5 x + 4 y)

\frac{d}{d x} (\frac{7}{x + 1})

\int_{1}^{5} w^{2} ln (w) d w

\int \frac{{ y }}{L { y }} d L

\frac{5}{( x ^{2} + 1 )}

What is the integral of 3xe^{(-2x)} ?
The integral of 3xe^{(-2x)} is 3/4 (-2e^{-2x}x-e^{-2x})+C