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} \cdot 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

- \frac{3 x ^{2}}{4 y ^{2}}

2 x^{2} + 15

\frac{d}{d x} (\frac{1}{- 3 + ( 3 x + 1 ) ^{2}})

y = sec (x), (\frac{π}{6}, 2 \frac{3}{3})

\frac{d y}{d x} = 1 + 8 y

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