What is the integral of (e^{6x})/(e^{2x)} ?

The integral of (e^{6x})/(e^{2x)} is 1/4 e^{4x}+C

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integral of (e^{6x})/(e^{2x)}

\int \frac{e ^{6 x}}{e ^{2 x}} d x

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

Solution steps

\int \frac{e ^{6 x}}{e ^{2 x}} d x

Simplify

\frac{e ^{6 x}}{e ^{2 x}} : e^{4 x}

= \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} (2 x + y^{10})

\int (π^{2} - 1) d x

\int 4 x ln (6 x) d x

100 y^{^{''}} - 20 y^{^{'}} + y = 0

(x^{2} y + 3 y) y^{^{'}} = 2 x y^{2} + 8 x

What is the integral of (e^{6x})/(e^{2x)} ?
The integral of (e^{6x})/(e^{2x)} is 1/4 e^{4x}+C