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

Simplify

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

Apply u-substitution

: \int - e^{u} d u

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

Take the constant out:

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

= - (- \int e^{u} d u)

Use the common integral:

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

= - (- e^{u})

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

x^{^{''}} - 5 x^{^{'}} + 6 x = cos (t) \cdot e^{- t}

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

\frac{x d y}{d x} + 3 y = x^{3} - x

y^{^{'}} = 2 y^{2} + x y^{2}, y (0) = 1

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