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

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

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

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

- \frac{1}{3} e^{- x^{3}} + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

= - \frac{1}{3} e^{u}

Substitute back

u = - x^{3}

= - \frac{1}{3} e^{- x^{3}}

Add a constant to the solution

= - \frac{1}{3} e^{- x^{3}} + C

y^{^{'}} - 2 y = 1 - 3 x

\frac{\partial}{\partial v} (\frac{u}{v})

y \cdot ln (x) - x^{2} y = x y^{^{'}}

\frac{\partial}{\partial x} (ln (x + y^{4}) - z)

\int (3 x^{2} - 2 x + 3) d x

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