What is the integral of x^3e^{-4x^2} ?

The integral of x^3e^{-4x^2} is 1/32 (-4e^{-4x^2}x^2-e^{-4x^2})+C

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integral of x^3e^{-4x^2}

\int x^{3} e^{- 4 x^{2}} d x

\frac{1}{32} (- 4 e^{- 4 x^{2}} x^{2} - e^{- 4 x^{2}}) + C

Solution steps

\int x^{3} e^{- 4 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{- 4 u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3} d u

= \int \frac{e ^{- 4 u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{- 4 u^{\frac{2}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int \frac{3 e ^{- 4 v} v}{2} d v

= \frac{1}{3} \cdot \int \frac{3 e ^{- 4 v} v}{2} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int e^{- 4 v} v d v

Apply u-substitution

: \int \frac{e ^{w} w}{16} d w

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int \frac{e ^{w} w}{16} d w

Take the constant out:

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

= \frac{1}{3} \cdot \frac{3}{2} \cdot \frac{1}{16} \cdot \int e^{w} w d w

Apply Integration By Parts

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

= \frac{1}{3} \cdot \frac{3}{2} \cdot \frac{1}{16} (e^{w} w - \int e^{w} d w)

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

= \frac{1}{3} \cdot \frac{3}{2} \cdot \frac{1}{16} (e^{w} w - e^{w})

Substitute back

= \frac{1}{3} \cdot \frac{3}{2} \cdot \frac{1}{16} (e^{- 4 (x^{3})^{\frac{2}{3}}} (- 4 (x^{3})^{\frac{2}{3}}) - e^{- 4 (x^{3})^{\frac{2}{3}}})

Simplify

\frac{1}{3} \cdot \frac{3}{2} \cdot \frac{1}{16} (e^{- 4 (x^{3})^{\frac{2}{3}}} (- 4 (x^{3})^{\frac{2}{3}}) - e^{- 4 (x^{3})^{\frac{2}{3}}}) : \frac{1}{32} (- 4 e^{- 4 x^{2}} x^{2} - e^{- 4 x^{2}})

= \frac{1}{32} (- 4 e^{- 4 x^{2}} x^{2} - e^{- 4 x^{2}})

Add a constant to the solution

= \frac{1}{32} (- 4 e^{- 4 x^{2}} x^{2} - e^{- 4 x^{2}}) + C

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What is the integral of x^3e^{-4x^2} ?
The integral of x^3e^{-4x^2} is 1/32 (-4e^{-4x^2}x^2-e^{-4x^2})+C