What is the integral of-(e^{2x})/(8+e^x) ?

The integral of-(e^{2x})/(8+e^x) is -8-e^x+8ln|8+e^x|+C

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integral of-(e^{2x})/(8+e^x)

\int - \frac{e ^{2 x}}{8 + e ^{x}} d x

- 8 - e^{x} + 8 ln ∣ 8 + e^{x} ∣ + C

Solution steps

\int - \frac{e ^{2 x}}{8 + e ^{x}} d x

Take the constant out:

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

= - \int \frac{e ^{2 x}}{8 + e ^{x}} d x

Apply u-substitution

: \int \frac{u - 8}{u} d u

= - \int \frac{u - 8}{u} d u

Expand

\frac{u - 8}{u} : 1 - \frac{8}{u}

= - \int 1 - \frac{8}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int 1 d u - \int \frac{8}{u} d u)

\int 1 d u = u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= - (u - 8 ln ∣ u ∣)

Substitute back

u = 8 + e^{x}

= - (8 + e^{x} - 8 ln ∣ 8 + e^{x} ∣)

Simplify

= - 8 - e^{x} + 8 ln ∣ 8 + e^{x} ∣

Add a constant to the solution

= - 8 - e^{x} + 8 ln ∣ 8 + e^{x} ∣ + C

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What is the integral of-(e^{2x})/(8+e^x) ?
The integral of-(e^{2x})/(8+e^x) is -8-e^x+8ln|8+e^x|+C