What is the integral of e^{-4x}x ?

The integral of e^{-4x}x is 1/16 (-4e^{-4x}x-e^{-4x})+C

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integral of e^{-4x}x

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

\frac{1}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

Solution steps

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

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - 4 x

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

Simplify

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

Add a constant to the solution

= \frac{1}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

\int \frac{2 x + 3}{2 x + 1} d x

\frac{\partial}{\partial y} (e^{x + y})

x \to 2 lim (9 - x^{2})

n = 1 \sum \infty \frac{7}{n ^{2} + 4}

x \to 0 lim (\frac{9 ^{x} - 7 ^{x}}{x})

What is the integral of e^{-4x}x ?
The integral of e^{-4x}x is 1/16 (-4e^{-4x}x-e^{-4x})+C