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

The integral of e^{-x^2}x^3 is 1/2 (-e^{-x^2}x^2-e^{-x^2})+C

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integral of e^{-x^2}x^3

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

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

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

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

Substitute back

u = - x^{2}

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

Simplify

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

Add a constant to the solution

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

(0, 14), (1, 9)

\int e^{- 5 t} d t

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

\int \frac{x ^{2} - 9}{x - 3} d x

\frac{d}{d x} ((x^{3} + 1)^{\frac{1}{3}})

What is the integral of e^{-x^2}x^3 ?
The integral of e^{-x^2}x^3 is 1/2 (-e^{-x^2}x^2-e^{-x^2})+C