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

The integral of x^9e^{x^5} is 1/5 (e^{x^5}x^5-e^{x^5})+C

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

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

\frac{1}{5} (e^{x^{5}} x^{5} - e^{x^{5}}) + C

Solution steps

\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

= \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

= \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

= \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

= \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

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

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

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

Substitute back

= \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

\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{1}{5} (e^{x^{5}} x^{5} - e^{x^{5}})

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

Add a constant to the solution

= \frac{1}{5} (e^{x^{5}} x^{5} - e^{x^{5}}) + C

\frac{d}{d x} (\frac{a + b x + c x ^{2}}{x})

\frac{d}{d x} (4 - cos (x))

x \to 2 + lim (4 - x^{2})

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

(6.3) (9.6)

What is the integral of x^9e^{x^5} ?
The integral of x^9e^{x^5} is 1/5 (e^{x^5}x^5-e^{x^5})+C