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

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

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

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

- 6 - e^{x} + 6 ln ∣ 6 + e^{x} ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 6}{u} d u

= - \int \frac{u - 6}{u} d u

Expand

\frac{u - 6}{u} : 1 - \frac{6}{u}

= - \int 1 - \frac{6}{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{6}{u} d u)

\int 1 d u = u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

= - (u - 6 ln ∣ u ∣)

Substitute back

u = 6 + e^{x}

= - (6 + e^{x} - 6 ln ∣ 6 + e^{x} ∣)

Simplify

= - 6 - e^{x} + 6 ln ∣ 6 + e^{x} ∣

Add a constant to the solution

= - 6 - e^{x} + 6 ln ∣ 6 + e^{x} ∣ + C

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