What is the integral of (e^{8x})/(e^{5x)} ?

The integral of (e^{8x})/(e^{5x)} is 1/3 e^{3x}+C

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integral of (e^{8x})/(e^{5x)}

\int \frac{e ^{8 x}}{e ^{5 x}} d x

\frac{1}{3} e^{3 x} + C

Solution steps

\int \frac{e ^{8 x}}{e ^{5 x}} d x

Simplify

\frac{e ^{8 x}}{e ^{5 x}} : e^{3 x}

= \int e^{3 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

= \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} e^{u}

Substitute back

u = 3 x

= \frac{1}{3} e^{3 x}

Add a constant to the solution

= \frac{1}{3} e^{3 x} + C

(0, - 6), (5, - 3)

\frac{d}{d x} (x cos (x y))

\int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

\frac{1}{1 - 2 x}

\frac{d}{d x} (\frac{x}{2.4 ( c ^{2} + c x )} (c^{2} - x^{2}))

What is the integral of (e^{8x})/(e^{5x)} ?
The integral of (e^{8x})/(e^{5x)} is 1/3 e^{3x}+C