What is the integral of 5/(6e^{2x)} ?

The integral of 5/(6e^{2x)} is -5/12 e^{-2x}+C

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integral of 5/(6e^{2x)}

\int \frac{5}{6 e ^{2 x}} d x

- \frac{5}{12} e^{- 2 x} + C

Solution steps

\int \frac{5}{6 e ^{2 x}} d x

Take the constant out:

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

= \frac{5}{6} \cdot \int \frac{1}{e ^{2 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \frac{5}{6} \cdot \int e^{- 2 x} d x

Apply u-substitution

= \frac{5}{6} \cdot \int - \frac{1}{2} e^{u} d u

Take the constant out:

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

= \frac{5}{6} (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{5}{6} (- \frac{1}{2} e^{u})

Substitute back

u = - 2 x

= \frac{5}{6} (- \frac{1}{2} e^{- 2 x})

Simplify

\frac{5}{6} (- \frac{1}{2} e^{- 2 x}) : - \frac{5}{12} e^{- 2 x}

= - \frac{5}{12} e^{- 2 x}

Add a constant to the solution

= - \frac{5}{12} e^{- 2 x} + C

\int \frac{x + 2}{( x + 1 ) ^{2} ( 2 x - 1 )} d x

\int \frac{sec ^{2} ( y )}{csc ( y )} d y

\frac{d}{d x} (\frac{3 x}{x ^{2} + 9})

x \to 0 lim (x^{2} - 2 x^{3})

n = 0 \sum \infty \frac{1}{n !} x^{n}

What is the integral of 5/(6e^{2x)} ?
The integral of 5/(6e^{2x)} is -5/12 e^{-2x}+C