What is the integral of-2e^{8x} ?

The integral of-2e^{8x} is -1/4 e^{8x}+C

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integral of-2e^{8x}

\int - 2 e^{8 x} d x

- \frac{1}{4} e^{8 x} + C

Solution steps

\int - 2 e^{8 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^{8 x} d x

Apply u-substitution

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

= - 2 \cdot \int 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

= - 2 \cdot \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= - 2 \cdot \frac{1}{8} e^{u}

Substitute back

u = 8 x

= - 2 \cdot \frac{1}{8} e^{8 x}

Simplify

- 2 \cdot \frac{1}{8} e^{8 x} : - \frac{1}{4} e^{8 x}

= - \frac{1}{4} e^{8 x}

Add a constant to the solution

= - \frac{1}{4} e^{8 x} + C

\frac{d}{d x} (ln^{2} (x + 3))

\frac{1}{x y}

f^{^{'}} (θ) = θ cos (θ) sin (θ)

y^{^{''}} + 8 y^{^{'}} + 32 y = 128, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d t} (x y (t) z (t))