What is the integral of 8e^{8x-3} ?

The integral of 8e^{8x-3} is e^{8x-3}+C

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integral of 8e^{8x-3}

\int 8 e^{8 x - 3} d x

e^{8 x - 3} + C

Solution steps

\int 8 e^{8 x - 3} d x

Take the constant out:

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

= 8 \cdot \int e^{8 x - 3} d x

Apply u-substitution

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

= 8 \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

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

Use the common integral:

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

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

Substitute back

u = 8 x - 3

= 8 \cdot \frac{1}{8} e^{8 x - 3}

Simplify

8 \cdot \frac{1}{8} e^{8 x - 3} : e^{8 x - 3}

= e^{8 x - 3}

Add a constant to the solution

= e^{8 x - 3} + C

x \to 6 lim ((\frac{1}{x - 6})^{2})

\frac{d}{d x} (x 3)

y = \frac{25 - x ^{2}}{4 x}

x y^{2} + 3 y^{2} - x^{2} \cdot \frac{d y}{d x} = 0

y^{^{'}} - y = t e^{3 t}