What is the integral of 2x^9e^{-x^5} ?

The integral of 2x^9e^{-x^5} is 2/5 (-e^{-x^5}x^5-e^{-x^5})+C

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integral of 2x^9e^{-x^5}

\int 2 x^{9} e^{- x^{5}} d x

\frac{2}{5} (- e^{- x^{5}} x^{5} - e^{- x^{5}}) + C

Solution steps

\int 2 x^{9} e^{- x^{5}} d x

Take the constant out:

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

= 2 \cdot \int x^{9} e^{- x^{5}} d x

Apply u-substitution

: \int \frac{e ^{- u^{\frac{5}{9}}} u ^{\frac{1}{9}}}{9} d u

= 2 \cdot \int \frac{e ^{- u^{\frac{5}{9}}} u ^{\frac{1}{9}}}{9} d u

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int e^{- u^{\frac{5}{9}}} u^{\frac{1}{9}} d u

Apply u-substitution

: \int \frac{9 e ^{- v} v}{5} d v

= 2 \cdot \frac{1}{9} \cdot \int \frac{9 e ^{- v} v}{5} d v

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \frac{9}{5} \cdot \int e^{- v} v d v

Apply Integration By Parts

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

= 2 \cdot \frac{1}{9} \cdot \frac{9}{5} (- e^{- v} v - \int - e^{- v} d v)

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

= 2 \cdot \frac{1}{9} \cdot \frac{9}{5} (- e^{- v} v - e^{- v})

Substitute back

= 2 \cdot \frac{1}{9} \cdot \frac{9}{5} (- e^{- (x^{9})^{\frac{5}{9}}} (x^{9})^{\frac{5}{9}} - e^{- (x^{9})^{\frac{5}{9}}})

Simplify

2 \cdot \frac{1}{9} \cdot \frac{9}{5} (- e^{- (x^{9})^{\frac{5}{9}}} (x^{9})^{\frac{5}{9}} - e^{- (x^{9})^{\frac{5}{9}}}) : \frac{2}{5} (- e^{- x^{5}} x^{5} - e^{- x^{5}})

= \frac{2}{5} (- e^{- x^{5}} x^{5} - e^{- x^{5}})

Add a constant to the solution

= \frac{2}{5} (- e^{- x^{5}} x^{5} - e^{- x^{5}}) + C

\int \frac{9}{x ^{2} x ^{2} + 4} d x

\frac{\partial}{\partial x} (x^{2} + z sin (x yz))

\int \frac{( 1 - u ^{2} - u - u ^{3} )}{1 + u ^{2}} d u

x^{2} - 3

\frac{d}{d z} (ln (\frac{b ^{2} - z ^{2}}{b ^{2} + z ^{2}}))

What is the integral of 2x^9e^{-x^5} ?
The integral of 2x^9e^{-x^5} is 2/5 (-e^{-x^5}x^5-e^{-x^5})+C