What is the integral of e^{-xr^2}3r ?

The integral of e^{-xr^2}3r is -e^{-xr^2} 3/(2x)+C

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integral of e^{-xr^2}3r

\int e^{- x r^{2}} 3 r d r

- e^{- x r^{2}} \frac{3}{2 x} + C

Solution steps

\int e^{- x r^{2}} \cdot 3 r d r

Take the constant out:

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

= 3 \cdot \int e^{- x r^{2}} r d r

Apply u-substitution

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

= 3 \cdot \int - \frac{e ^{u}}{2 x} d u

Take the constant out:

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

= 3 (- \frac{1}{2 x} \cdot \int e^{u} d u)

Use the common integral:

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

= 3 (- \frac{1}{2 x} e^{u})

Substitute back

u = - x r^{2}

= 3 (- \frac{1}{2 x} e^{- x r^{2}})

Simplify

3 (- \frac{1}{2 x} e^{- x r^{2}}) : - e^{- x r^{2}} \frac{3}{2 x}

= - e^{- x r^{2}} \frac{3}{2 x}

Add a constant to the solution

= - e^{- x r^{2}} \frac{3}{2 x} + C

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What is the integral of e^{-xr^2}3r ?
The integral of e^{-xr^2}3r is -e^{-xr^2} 3/(2x)+C