What is the integral of 3x*e^{(-4x)} ?

The integral of 3x*e^{(-4x)} is 3/16 (-4e^{-4x}x-e^{-4x})+C

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integral of 3x*e^{(-4x)}

\int 3 x \cdot e^{(- 4 x)} d x

\frac{3}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

Solution steps

\int 3 x e^{- 4 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^{- 4 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= 3 \cdot \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= 3 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= 3 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = - 4 x

= 3 \cdot \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x})

Simplify

3 \cdot \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x}) : \frac{3}{16} (- 4 e^{- 4 x} x - e^{- 4 x})

= \frac{3}{16} (- 4 e^{- 4 x} x - e^{- 4 x})

Add a constant to the solution

= \frac{3}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

\frac{d}{d x} (\frac{x ^{2} y}{x ^{2} + y ^{2}})

y = x^{2} - 8

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x^{2}

\int \frac{x}{2 x} d x

x \to \infty lim (\frac{x}{k})

What is the integral of 3x*e^{(-4x)} ?
The integral of 3x*e^{(-4x)} is 3/16 (-4e^{-4x}x-e^{-4x})+C