What is the integral of 9xe^{-4x^2} ?

The integral of 9xe^{-4x^2} is -9/8 e^{-4x^2}+C

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integral of 9xe^{-4x^2}

\int 9 x e^{- 4 x^{2}} d x

- \frac{9}{8} e^{- 4 x^{2}} + C

Solution steps

\int 9 x e^{- 4 x^{2}} d x

Take the constant out:

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

= 9 \cdot \int x e^{- 4 x^{2}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{8} d u

= 9 \cdot \int - \frac{e ^{u}}{8} d u

Take the constant out:

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

= 9 (- \frac{1}{8} \cdot \int e^{u} d u)

Use the common integral:

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

= 9 (- \frac{1}{8} e^{u})

Substitute back

u = - 4 x^{2}

= 9 (- \frac{1}{8} e^{- 4 x^{2}})

Simplify

9 (- \frac{1}{8} e^{- 4 x^{2}}) : - \frac{9}{8} e^{- 4 x^{2}}

= - \frac{9}{8} e^{- 4 x^{2}}

Add a constant to the solution

= - \frac{9}{8} e^{- 4 x^{2}} + C

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What is the integral of 9xe^{-4x^2} ?
The integral of 9xe^{-4x^2} is -9/8 e^{-4x^2}+C