What is the integral of e^{3x}(e^{3x-8})^5 ?

The integral of e^{3x}(e^{3x-8})^5 is 1/18 e^{18x-40}+C

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integral of e^{3x}(e^{3x-8})^5

\int e^{3 x} (e^{3 x - 8})^{5} d x

\frac{1}{18} e^{18 x - 40} + C

Solution steps

\int e^{3 x} (e^{3 x - 8})^{5} d x

Simplify

e^{3 x} (e^{3 x - 8})^{5} : e^{18 x - 40}

= \int e^{18 x - 40} d x

Apply u-substitution

: \int e^{u} \frac{1}{18} d u

= \int e^{u} \frac{1}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{18} e^{u}

Substitute back

u = 18 x - 40

= \frac{1}{18} e^{18 x - 40}

Add a constant to the solution

= \frac{1}{18} e^{18 x - 40} + C

y^{^{'}} = 3 x + y

\frac{d ^{3}}{d x ^{3}} (ln (x + 1))

\frac{d y}{d x}, y^{3} = x + y

\int x + \frac{1}{x} d x

\frac{\partial}{\partial y} (\frac{x + 2 y}{x ^{2} - y})

What is the integral of e^{3x}(e^{3x-8})^5 ?
The integral of e^{3x}(e^{3x-8})^5 is 1/18 e^{18x-40}+C