What is the integral of 2x*e^{-2x} ?

The integral of 2x*e^{-2x} is -e^{-2x}x-1/2 e^{-2x}+C

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integral of 2x*e^{-2x}

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

- e^{- 2 x} x - \frac{1}{2} e^{- 2 x} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - 2 x

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

Simplify

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

= - e^{- 2 x} x - \frac{1}{2} e^{- 2 x}

Add a constant to the solution

= - e^{- 2 x} x - \frac{1}{2} e^{- 2 x} + C

\int \frac{x ^{3} - 6 x ^{2} + 4 x - 10}{x ^{2} - 1} d x

\int cos (4 x) cos (7 x) d x

f (x) = - \frac{6}{x ^{4}}

x \to \frac{π}{4} lim (cos (x))

n \to \infty lim (n 10 n)

What is the integral of 2x*e^{-2x} ?
The integral of 2x*e^{-2x} is -e^{-2x}x-1/2 e^{-2x}+C