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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int e^{v} d v

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

Use the common integral:

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

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

Substitute back

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

Add a constant to the solution

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

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

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

\frac{d}{d x} (x^{3} (x - 8)^{2})

\frac{d}{d x} (\frac{1}{x - y})

\frac{d y}{d x} + \frac{1}{x} y = 0

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