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

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

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integral of-2xe^{-x^2}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - 2 \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

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

Use the common integral:

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

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

Substitute back

u = - x^{2}

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

Simplify

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

= e^{- x^{2}}

Add a constant to the solution

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

\int_{1}^{4} \frac{x ^{2} + 7}{5 x - x ^{2}} d x

y^{^{'}} = e^{- y} (x^{2} - 6)

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 3

\int e^{x} 36 - e^{2 x} d x

\frac{4 e ^{x}}{cos ( x )}