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

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

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integral of r*e^{-r^2}

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

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

Solution steps

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

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 = - r^{2}

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

Add a constant to the solution

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

\int \frac{16 - x ^{2}}{x} d x

\int_{- 1}^{0} 8 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

y^{^{'}} = - \frac{x y}{x ^{2} + y ^{2}}

\int \frac{x ^{2}}{1 + x ^{2}} d x

x y^{^{'}} + 4 y = 8 x^{4}, y (1) = 2

What is the integral of r*e^{-r^2} ?
The integral of r*e^{-r^2} is -1/2 e^{-r^2}+C