What is the integral of x^9*e^{x^{10}-9} ?

The integral of x^9*e^{x^{10}-9} is 1/10 e^{x^{10}-9}+C

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integral of x^9*e^{x^{10}-9}

\int x^{9} \cdot e^{x^{10} - 9} d x

\frac{1}{10} e^{x^{10} - 9} + C

Solution steps

\int x^{9} e^{x^{10} - 9} d x

x^{9} e^{x^{10} - 9} = e^{- 9} x^{9} e^{x^{10}}

= \int e^{- 9} x^{9} e^{x^{10}} d x

Take the constant out:

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

= e^{- 9} \cdot \int x^{9} e^{x^{10}} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{9}} \cdot \int x^{9} e^{x^{10}} d x

Apply u-substitution

: \int \frac{e ^{u^{\frac{10}{9}}} u ^{\frac{1}{9}}}{9} d u

= \frac{1}{e ^{9}} \cdot \int \frac{e ^{u^{\frac{10}{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}{e ^{9}} \cdot \frac{1}{9} \cdot \int e^{u^{\frac{10}{9}}} u^{\frac{1}{9}} d u

Apply u-substitution

: \int \frac{9 e ^{v}}{10} d v

= \frac{1}{e ^{9}} \cdot \frac{1}{9} \cdot \int \frac{9 e ^{v}}{10} d v

Take the constant out:

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

= \frac{1}{e ^{9}} \cdot \frac{1}{9} \cdot \frac{9}{10} \cdot \int e^{v} d v

Use the common integral:

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

= \frac{1}{e ^{9}} \cdot \frac{1}{9} \cdot \frac{9}{10} e^{v}

Substitute back

= \frac{1}{e ^{9}} \cdot \frac{1}{9} \cdot \frac{9}{10} e^{(x^{9})^{\frac{10}{9}}}

Simplify

\frac{1}{e ^{9}} \cdot \frac{1}{9} \cdot \frac{9}{10} e^{(x^{9})^{\frac{10}{9}}} : \frac{1}{10} e^{x^{10} - 9}

= \frac{1}{10} e^{x^{10} - 9}

Add a constant to the solution

= \frac{1}{10} e^{x^{10} - 9} + C

y = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

n = 0 \sum \infty e^{- \frac{9 n}{2}}

\int u^{- 3} d u

\frac{d}{d t} (t^{\frac{3}{2}})

\frac{d y}{d x} = (x - 3) e^{- 2 y}

What is the integral of x^9*e^{x^{10}-9} ?
The integral of x^9*e^{x^{10}-9} is 1/10 e^{x^{10}-9}+C