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 \frac{1}{( x ^{2} + 4 x + 5 ) ^{2}} d x

\int \frac{1}{( x + 4 ) ( x + 8 )} d x

(3, - 2), (7, 0)

x \to 1 lim (\frac{x - 2}{x ^{2} - 1})

(3.12) (5.6)