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

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

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integral of (-4x)/(e^{x^2)}

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

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

Solution steps

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

Take the constant out:

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

= - 4 \cdot \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}}

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

Apply u-substitution

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

= - 4 \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

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

Use the common integral:

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

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

Substitute back

u = - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d y} (3 x e^{- \frac{1}{y ^{3}}} - 20 x^{3})

f (x) = x - \frac{1}{x} + ln (x), at x = 1

y = 16 x sin (x), (\frac{π}{2}, 8 π)

\frac{\partial}{\partial y} (x ln (y + z))

\int \frac{3}{x - 3} d x

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