What is the integral of 16xe^{-x^2} ?

The integral of 16xe^{-x^2} is -8e^{-x^2}+C

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integral of 16xe^{-x^2}

\int 16 x e^{- x^{2}} d x

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

Solution steps

\int 16 x e^{- x^{2}} d x

Take the constant out:

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

= 16 \cdot \int x e^{- x^{2}} d x

Apply u-substitution

= 16 \cdot \int - \frac{e ^{u}}{2} d u

Take the constant out:

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

= 16 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= 16 (- \frac{1}{2} e^{u})

Substitute back

u = - x^{2}

= 16 (- \frac{1}{2} e^{- x^{2}})

Simplify

16 (- \frac{1}{2} e^{- x^{2}}) : - 8 e^{- x^{2}}

= - 8 e^{- x^{2}}

Add a constant to the solution

= - 8 e^{- x^{2}} + C

What is the integral of 16xe^{-x^2} ?
The integral of 16xe^{-x^2} is -8e^{-x^2}+C