What is the integral of xe^{-1/10 x^2} ?

The integral of xe^{-1/10 x^2} is -5e^{-(x^2)/(10)}+C

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integral of xe^{-1/10 x^2}

\int x e^{- \frac{1}{10} x^{2}} d x

- 5 e^{- \frac{x ^{2}}{10}} + C

Solution steps

\int x e^{- \frac{1}{10} x^{2}} d x

Simplify

- \frac{1}{10} x^{2} : - \frac{x ^{2}}{10}

= \int x e^{- \frac{x ^{2}}{10}} d x

Apply u-substitution

: \int - 5 e^{u} d u

= \int - 5 e^{u} d u

Take the constant out:

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

= - 5 \cdot \int e^{u} d u

Use the common integral:

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

= - 5 e^{u}

Substitute back

u = - \frac{x ^{2}}{10}

= - 5 e^{- \frac{x ^{2}}{10}}

Add a constant to the solution

= - 5 e^{- \frac{x ^{2}}{10}} + C

\frac{d ^{3}}{d x ^{3}} ((1 + x)^{3})

\frac{\partial}{\partial x} (2 x y^{3} + 2)

y^{^{'}} - (\frac{3}{t}) \cdot y = 2 t^{3} \cdot e^{(2 t)}, y (1) = 0

\frac{d}{d x} (ln (1 - e^{- x}))

n = 1 \sum \infty \frac{1}{2 ^{n} - 5}

What is the integral of xe^{-1/10 x^2} ?
The integral of xe^{-1/10 x^2} is -5e^{-(x^2)/(10)}+C