What is the integral of 1/6 (e^{6x}) ?

The integral of 1/6 (e^{6x}) is 1/36 e^{6x}+C

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integral of 1/6 (e^{6x})

\int \frac{1}{6} (e^{6 x}) d x

\frac{1}{36} e^{6 x} + C

Solution steps

\int \frac{1}{6} e^{6 x} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int e^{6 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{6} d u

= \frac{1}{6} \cdot \int e^{u} \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{6} \cdot \frac{1}{6} e^{u}

Substitute back

u = 6 x

= \frac{1}{6} \cdot \frac{1}{6} e^{6 x}

Simplify

\frac{1}{6} \cdot \frac{1}{6} e^{6 x} : \frac{1}{36} e^{6 x}

= \frac{1}{36} e^{6 x}

Add a constant to the solution

= \frac{1}{36} e^{6 x} + C

y = cos (e^{x})

n = 0 \sum \infty \frac{n ^{5}}{2 ^{n}}

f (x) = \frac{8}{x ^{2} + 4}, (2, 1)

x \to \infty lim (2)

\int_{0}^{\infty} 8 x e^{2 x} d x

What is the integral of 1/6 (e^{6x}) ?
The integral of 1/6 (e^{6x}) is 1/36 e^{6x}+C