What is the integral of 4xe^{-5x} ?

The integral of 4xe^{-5x} is 4/25 (-5e^{-5x}x-e^{-5x})+C

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integral of 4xe^{-5x}

\int 4 x e^{- 5 x} d x

\frac{4}{25} (- 5 e^{- 5 x} x - e^{- 5 x}) + C

Solution steps

\int 4 x e^{- 5 x} d x

Take the constant out:

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

= 4 \cdot \int x e^{- 5 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{25} d u

= 4 \cdot \int \frac{e ^{u} u}{25} d u

Take the constant out:

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

= 4 \cdot \frac{1}{25} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 4 \cdot \frac{1}{25} (e^{u} u - \int e^{u} d u)

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

= 4 \cdot \frac{1}{25} (e^{u} u - e^{u})

Substitute back

u = - 5 x

= 4 \cdot \frac{1}{25} (e^{- 5 x} (- 5 x) - e^{- 5 x})

Simplify

4 \cdot \frac{1}{25} (e^{- 5 x} (- 5 x) - e^{- 5 x}) : \frac{4}{25} (- 5 e^{- 5 x} x - e^{- 5 x})

= \frac{4}{25} (- 5 e^{- 5 x} x - e^{- 5 x})

Add a constant to the solution

= \frac{4}{25} (- 5 e^{- 5 x} x - e^{- 5 x}) + C

x \to a lim (\frac{1}{x})

\int sin (5 t) d t

x \to - 5 lim (\frac{1}{x + 5})

\frac{d}{d x} (\frac{3}{5} x^{15})

(3 x + 5 y) d x + (5 x - 8 y 3) d y = 0

What is the integral of 4xe^{-5x} ?
The integral of 4xe^{-5x} is 4/25 (-5e^{-5x}x-e^{-5x})+C