What is the integral of e^{-1/4 x^2}x ?

The integral of e^{-1/4 x^2}x is -2e^{-(x^2)/4}+C

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integral of e^{-1/4 x^2}x

\int e^{- \frac{1}{4} x^{2}} x d x

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

Solution steps

\int e^{- \frac{1}{4} x^{2}} x d x

Simplify

- \frac{1}{4} x^{2} : - \frac{x ^{2}}{4}

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

Apply u-substitution

: \int - 2 e^{u} d u

= \int - 2 e^{u} d u

Take the constant out:

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

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

Use the common integral:

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

= - 2 e^{u}

Substitute back

u = - \frac{x ^{2}}{4}

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

Add a constant to the solution

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

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What is the integral of e^{-1/4 x^2}x ?
The integral of e^{-1/4 x^2}x is -2e^{-(x^2)/4}+C