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

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

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

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

\frac{1}{2} ln e^{2 x} + 6 + C

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = e^{2 x} + 6

= \frac{1}{2} ln e^{2 x} + 6

Add a constant to the solution

= \frac{1}{2} ln e^{2 x} + 6 + C

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

\frac{d}{d x} (r (x) cos^{2} (r (x) x))

\int e^{8 x} cos (9 x) d x

\int I n (4 x) d x

f (x) = \frac{2}{x + 1}, at x = 3, y = 1

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