What is the integral of x^8e^{x^9+8} ?

The integral of x^8e^{x^9+8} is 1/9 e^{8+x^9}+C

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integral of x^8e^{x^9+8}

\int x^{8} e^{x^{9} + 8} d x

\frac{1}{9} e^{8 + x^{9}} + C

Solution steps

\int x^{8} e^{x^{9} + 8} d x

x^{8} e^{x^{9} + 8} = e^{8} x^{8} e^{x^{9}}

= \int e^{8} x^{8} e^{x^{9}} d x

Take the constant out:

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

= e^{8} \cdot \int x^{8} e^{x^{9}} d x

Apply u-substitution

: \int \frac{e ^{u^{\frac{9}{8}}} u ^{\frac{1}{8}}}{8} d u

= e^{8} \cdot \int \frac{e ^{u^{\frac{9}{8}}} u ^{\frac{1}{8}}}{8} d u

Take the constant out:

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

= e^{8} \frac{1}{8} \cdot \int e^{u^{\frac{9}{8}}} u^{\frac{1}{8}} d u

Apply u-substitution

: \int \frac{8 e ^{v}}{9} d v

= e^{8} \frac{1}{8} \cdot \int \frac{8 e ^{v}}{9} d v

Take the constant out:

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

= e^{8} \frac{1}{8} \cdot \frac{8}{9} \cdot \int e^{v} d v

Use the common integral:

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

= e^{8} \frac{1}{8} \cdot \frac{8}{9} e^{v}

Substitute back

= e^{8} \frac{1}{8} \cdot \frac{8}{9} e^{(x^{8})^{\frac{9}{8}}}

Simplify

e^{8} \frac{1}{8} \cdot \frac{8}{9} e^{(x^{8})^{\frac{9}{8}}} : \frac{1}{9} e^{8 + x^{9}}

= \frac{1}{9} e^{8 + x^{9}}

Add a constant to the solution

= \frac{1}{9} e^{8 + x^{9}} + C

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What is the integral of x^8e^{x^9+8} ?
The integral of x^8e^{x^9+8} is 1/9 e^{8+x^9}+C