What is the integral of 7x^{17}e^{-x^9} ?

The integral of 7x^{17}e^{-x^9} is 7/9 (-e^{-x^9}x^9-e^{-x^9})+C

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integral of 7x^{17}e^{-x^9}

\int 7 x^{17} e^{- x^{9}} d x

\frac{7}{9} (- e^{- x^{9}} x^{9} - e^{- x^{9}}) + C

Solution steps

\int 7 x^{17} e^{- x^{9}} d x

Take the constant out:

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

= 7 \cdot \int x^{17} e^{- x^{9}} d x

Apply u-substitution

: \int \frac{e ^{- u^{\frac{9}{17}}} u ^{\frac{1}{17}}}{17} d u

= 7 \cdot \int \frac{e ^{- u^{\frac{9}{17}}} u ^{\frac{1}{17}}}{17} d u

Take the constant out:

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

= 7 \cdot \frac{1}{17} \cdot \int e^{- u^{\frac{9}{17}}} u^{\frac{1}{17}} d u

Apply u-substitution

: \int \frac{17 e ^{- v} v}{9} d v

= 7 \cdot \frac{1}{17} \cdot \int \frac{17 e ^{- v} v}{9} d v

Take the constant out:

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

= 7 \cdot \frac{1}{17} \cdot \frac{17}{9} \cdot \int e^{- v} v d v

Apply Integration By Parts

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

= 7 \cdot \frac{1}{17} \cdot \frac{17}{9} (- e^{- v} v - \int - e^{- v} d v)

\int - e^{- v} d v = e^{- v}

= 7 \cdot \frac{1}{17} \cdot \frac{17}{9} (- e^{- v} v - e^{- v})

Substitute back

= 7 \cdot \frac{1}{17} \cdot \frac{17}{9} (- e^{- (x^{17})^{\frac{9}{17}}} (x^{17})^{\frac{9}{17}} - e^{- (x^{17})^{\frac{9}{17}}})

Simplify

7 \cdot \frac{1}{17} \cdot \frac{17}{9} (- e^{- (x^{17})^{\frac{9}{17}}} (x^{17})^{\frac{9}{17}} - e^{- (x^{17})^{\frac{9}{17}}}) : \frac{7}{9} (- e^{- x^{9}} x^{9} - e^{- x^{9}})

= \frac{7}{9} (- e^{- x^{9}} x^{9} - e^{- x^{9}})

Add a constant to the solution

= \frac{7}{9} (- e^{- x^{9}} x^{9} - e^{- x^{9}}) + C

\frac{\partial}{\partial t} (e^{s t})

\frac{d}{d x} (a x sin (x))

\frac{d y}{d x} = x y cos (x)

(\frac{d y}{d t}) - (\frac{3}{t}) \cdot y = 2 t^{3} e^{2 t}, y (1) = 0

f (x) = 7 x^{2} + 6 x

What is the integral of 7x^{17}e^{-x^9} ?
The integral of 7x^{17}e^{-x^9} is 7/9 (-e^{-x^9}x^9-e^{-x^9})+C