What is the integral of (6x)/(e^{x^2)} ?

The integral of (6x)/(e^{x^2)} is -3e^{-x^2}+C

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integral of (6x)/(e^{x^2)}

\int \frac{6 x}{e ^{x^{2}}} d x

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

Solution steps

\int \frac{6 x}{e ^{x^{2}}} d x

Take the constant out:

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

= 6 \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}}

= 6 \cdot \int x e^{- x^{2}} d x

Apply u-substitution

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

= 6 \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

= 6 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= 6 (- \frac{1}{2} e^{u})

Substitute back

u = - x^{2}

= 6 (- \frac{1}{2} e^{- x^{2}})

Simplify

6 (- \frac{1}{2} e^{- x^{2}}) : - 3 e^{- x^{2}}

= - 3 e^{- x^{2}}

Add a constant to the solution

= - 3 e^{- x^{2}} + C

\frac{d}{d x} ((x^{3} - 3 x)^{4})

\int y sin (z) d x

\int ln (15 x) d x

y^{^{''}} - 20 y^{^{'}} + 100 y = 0

\frac{\partial}{\partial x} (y^{2 x})

What is the integral of (6x)/(e^{x^2)} ?
The integral of (6x)/(e^{x^2)} is -3e^{-x^2}+C