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

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

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

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

\frac{7}{4} (- e^{- x^{4}} x^{4} - e^{- x^{4}}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 7 \cdot \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

= 7 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 7 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= 7 \cdot \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = - x^{4}

= 7 \cdot \frac{1}{4} (e^{- x^{4}} (- x^{4}) - e^{- x^{4}})

Simplify

7 \cdot \frac{1}{4} (e^{- x^{4}} (- x^{4}) - e^{- x^{4}}) : \frac{7}{4} (- e^{- x^{4}} x^{4} - e^{- x^{4}})

= \frac{7}{4} (- e^{- x^{4}} x^{4} - e^{- x^{4}})

Add a constant to the solution

= \frac{7}{4} (- e^{- x^{4}} x^{4} - e^{- x^{4}}) + C

(ln (5 x))^{^{'}}

\int_{0}^{2} \frac{x}{1 + 2 x ^{2}} d x

\frac{d}{d x} (\frac{e ^{x}}{4 x ^{2}})

\frac{d y}{d x} = (x - 1) (1 - y)

y^{^{''}} - 8 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 9

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