What is the integral of (e^{4x})/4 ?

The integral of (e^{4x})/4 is 1/16 e^{4x}+C

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integral of (e^{4x})/4

\int \frac{e ^{4 x}}{4} d x

\frac{1}{16} e^{4 x} + C

Solution steps

\int \frac{e ^{4 x}}{4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{4 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

= \frac{1}{4} \cdot \int e^{u} \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{4} \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 x

= \frac{1}{4} \cdot \frac{1}{4} e^{4 x}

Simplify

\frac{1}{4} \cdot \frac{1}{4} e^{4 x} : \frac{1}{16} e^{4 x}

= \frac{1}{16} e^{4 x}

Add a constant to the solution

= \frac{1}{16} e^{4 x} + C

x \to \infty lim (n^{x})

x \to - 3 lim (\frac{- 6 x - 18}{x ^{3} + 3 x ^{2}})

n = 1 \sum \infty \frac{n !}{n ^{2}}

\int_{0}^{π} sin^{2} (9 x) d x

\frac{d}{d x} (- 2 x \cdot e^{- x^{2}})

What is the integral of (e^{4x})/4 ?
The integral of (e^{4x})/4 is 1/16 e^{4x}+C