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

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

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integral of (e^{4x})

\int (e^{4 x}) d x

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

Solution steps

\int e^{4 x} d x

Apply u-substitution

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

= \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 \int e^{u} d u

Use the common integral:

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

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

Substitute back

u = 4 x

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (+ 4 x - 5 + 6 x y)

\frac{d ^{2} x}{d t ^{2}} = 5 + 2 x

\frac{d N}{d t} + N = Nt e^{t + 3}

5 x + 6

x + y