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 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

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

Substitute back

u = x^{2}

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

Simplify

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

= - e^{x^{2}}

Add a constant to the solution

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

\frac{d}{d x} (\frac{2}{ln ( 10 ) x ^{3}})

\int_{4}^{7} \frac{1}{x ^{2} - 16} d x

\frac{d}{d x} (2 x)

f (x) = x^{2} - 2 x - 6, at x = - 2

x \to 8 lim (\frac{4}{x - 8})