What is the integral of (-x)/(e^{x^2)} ?

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

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integral of (-x)/(e^{x^2)}

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

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

= - \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

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

Use the common integral:

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

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

Substitute back

u = - x^{2}

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

Simplify

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

Add a constant to the solution

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

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