What is the integral of x^7e^{-x^4} ?

The integral of x^7e^{-x^4} is 1/4 (-e^{-x^4}x^4-e^{-x^4})+C

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integral of x^7e^{-x^4}

\int x^{7} e^{- x^{4}} d x

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

Solution steps

\int x^{7} e^{- x^{4}} d x

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

= \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

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

Apply Integration By Parts

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

= \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = - x^{4}

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

Simplify

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

Add a constant to the solution

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

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What is the integral of x^7e^{-x^4} ?
The integral of x^7e^{-x^4} is 1/4 (-e^{-x^4}x^4-e^{-x^4})+C