What is the integral of 2e^{4x} ?

The integral of 2e^{4x} is 1/2 e^{4x}+C

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integral of 2e^{4x}

\int 2 e^{4 x} d x

\frac{1}{2} e^{4 x} + C

Solution steps

\int 2 e^{4 x} d x

Take the constant out:

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

= 2 \cdot \int e^{4 x} d x

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 x

= 2 \cdot \frac{1}{4} e^{4 x}

Simplify

2 \cdot \frac{1}{4} e^{4 x} : \frac{1}{2} e^{4 x}

= \frac{1}{2} e^{4 x}

Add a constant to the solution

= \frac{1}{2} e^{4 x} + C

\frac{5}{3 ( s + 3 ) ^{2} + 27}

x \to - 1 lim (\frac{x}{x - 1})

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

x \to 0 lim (\frac{e ^{x} - 1}{x ^{9}})

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = 0