What is the integral of 4xe^{-2x} ?

The integral of 4xe^{-2x} is -2e^{-2x}x-e^{-2x}+C

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integral of 4xe^{-2x}

\int 4 x e^{- 2 x} d x

- 2 e^{- 2 x} x - e^{- 2 x} + C

Solution steps

\int 4 x e^{- 2 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^{- 2 x} d x

Apply u-substitution

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

= 4 \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - 2 x

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

Simplify

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

= - 2 e^{- 2 x} x - e^{- 2 x}

Add a constant to the solution

= - 2 e^{- 2 x} x - e^{- 2 x} + C

(x^{2} - 25 ln (x^{2}))^{^{'}}

\int \frac{10}{15 sin ( x ) - 8 cos ( x )} d x

\int \frac{6}{x 36 - ( ln ( x ) ) ^{2}} d x

f (x) = 6 x cos (x), (π, - 6 π)

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