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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-e^{x^2}x

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

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

Solution steps

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

Take the constant out:

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

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

\frac{d}{d x} (x^{l o g_{6} (x)})

\int - 6 cos (x) d x

\int 11 ln (x^{2} - x + 2) d x

\frac{( e ^{x} + e ^{- x} )}{2}

y^{^{'}} - 2 t y = 4 t^{3}