What is the integral of 5xe^{-2x} ?

The integral of 5xe^{-2x} is 5/4 (-2e^{-2x}x-e^{-2x})+C

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integral of 5xe^{-2x}

\int 5 x e^{- 2 x} d x

\frac{5}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + C

Solution steps

\int 5 x e^{- 2 x} d x

Take the constant out:

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

= 5 \cdot \int x e^{- 2 x} d x

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - 2 x

= 5 \cdot \frac{1}{4} (e^{- 2 x} (- 2 x) - e^{- 2 x})

Simplify

5 \cdot \frac{1}{4} (e^{- 2 x} (- 2 x) - e^{- 2 x}) : \frac{5}{4} (- 2 e^{- 2 x} x - e^{- 2 x})

= \frac{5}{4} (- 2 e^{- 2 x} x - e^{- 2 x})

Add a constant to the solution

= \frac{5}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + C

\frac{2}{s + 3}

\frac{s + 1}{( s ^{2} + 9 )}

f (x) = 2 x^{2} - 3 x - 2, at x = 6

\int_{- \infty}^{- 1} e^{- 4 t} d t

f (x) = (x^{2} - 4) (3 x + 1)

What is the integral of 5xe^{-2x} ?
The integral of 5xe^{-2x} is 5/4 (-2e^{-2x}x-e^{-2x})+C