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

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

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

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

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

Solution steps

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

x \to + 2 lim (3 x^{2} + 4 x - 3)

y = - x^{2} + 2 x + 5, y = \frac{x ^{2}}{2} - 4 x + 5

x \to 0 lim (ln (1 + \frac{1}{x}))

\int e^{- 7 t} d t

16 y^{^{''}} + 2 y = 0

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