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

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

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integral of-xe^{x^2}

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

- \frac{1}{2} e^{x^{2}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - \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

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

Use the common integral:

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

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

Substitute back

u = x^{2}

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

Add a constant to the solution

= - \frac{1}{2} e^{x^{2}} + C

x = sec (\frac{1}{y})

\int - 4 x^{3} d x

\int_{0}^{1} (6 x^{2} - 1) e^{x} d x

x \to - 1 lim (\frac{1}{x ^{2}} + 3 x - 1)

\frac{d y}{d x} = (y + 1)^{(- 1)}