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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply u-substitution

: \int e^{v} \cdot 2 d v

= - \frac{1}{2} \cdot \int e^{v} \cdot 2 d v

Take the constant out:

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

= - \frac{1}{2} \cdot 2 \cdot \int e^{v} d v

Use the common integral:

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

= - \frac{1}{2} \cdot 2 e^{v}

Substitute back

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

Simplify

- \frac{1}{2} \cdot 2 e^{\frac{x ^{2}}{2}} : - e^{\frac{x ^{2}}{2}}

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

Add a constant to the solution

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

x \to 6 lim (x - 6)

y = 3 x^{\frac{2}{3}}

\frac{1}{x ^{0.5}}

\frac{\partial}{\partial y} ((5 x^{3} y + 4 y^{5})^{3})

x^{2} + 3 x - 2

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