What is the integral of (2e^{2x})/(e^{10)-1} ?

The integral of (2e^{2x})/(e^{10)-1} is 1/(e^{10)-1}e^{2x}+C

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integral of (2e^{2x})/(e^{10)-1}

\int \frac{2 e ^{2 x}}{e ^{10} - 1} d x

\frac{1}{e ^{10} - 1} e^{2 x} + C

Solution steps

\int \frac{2 e ^{2 x}}{e ^{10} - 1} d x

Take the constant out:

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

= \frac{2}{e ^{10} - 1} \cdot \int e^{2 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= \frac{2}{e ^{10} - 1} \cdot \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{2}{e ^{10} - 1} \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{2}{e ^{10} - 1} \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x

= \frac{2}{e ^{10} - 1} \cdot \frac{1}{2} e^{2 x}

Simplify

\frac{2}{e ^{10} - 1} \cdot \frac{1}{2} e^{2 x} : \frac{1}{e ^{10} - 1} e^{2 x}

= \frac{1}{e ^{10} - 1} e^{2 x}

Add a constant to the solution

= \frac{1}{e ^{10} - 1} e^{2 x} + C

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x \to 0 + lim (\frac{x - 2}{x})

What is the integral of (2e^{2x})/(e^{10)-1} ?
The integral of (2e^{2x})/(e^{10)-1} is 1/(e^{10)-1}e^{2x}+C