What is the integral of 4/(e^{2x)} ?

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

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integral of 4/(e^{2x)}

\int \frac{4}{e ^{2 x}} d x

- 2 e^{- 2 x} + C

Solution steps

\int \frac{4}{e ^{2 x}} d x

Take the constant out:

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

= 4 \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}

= 4 \cdot \int e^{- 2 x} d x

Apply u-substitution

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

= 4 \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

= 4 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= 4 (- \frac{1}{2} e^{u})

Substitute back

u = - 2 x

= 4 (- \frac{1}{2} e^{- 2 x})

Simplify

4 (- \frac{1}{2} e^{- 2 x}) : - 2 e^{- 2 x}

= - 2 e^{- 2 x}

Add a constant to the solution

= - 2 e^{- 2 x} + C

\int \frac{( 1 - 3 x )}{( 3 + 2 x )} d x

\frac{y ^{^{'}} + y ^{2}}{x ^{2}} = 0

\frac{d}{d M} (M^{2} (\frac{C}{2} - \frac{M}{3}))

x \to 0 lim (e^{1 - \frac{1}{x}})

\frac{d}{d x} (x (2 x + 1)^{4})