What is the integral of x^5e^{-x^3} ?

The integral of x^5e^{-x^3} is 1/3 (-e^{-x^3}x^3-e^{-x^3})+C

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integral of x^5e^{-x^3}

\int x^{5} e^{- x^{3}} d x

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

Solution steps

\int x^{5} e^{- x^{3}} d x

Apply u-substitution

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

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

Apply Integration By Parts

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

= \frac{1}{3} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - x^{3}

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

Simplify

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

Add a constant to the solution

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

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What is the integral of x^5e^{-x^3} ?
The integral of x^5e^{-x^3} is 1/3 (-e^{-x^3}x^3-e^{-x^3})+C