What is the integral of-(e^{2x})/(1+e^x) ?

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

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integral of-(e^{2x})/(1+e^x)

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

- 1 - e^{x} + ln ∣ 1 + e^{x} ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - (u - ln ∣ u ∣)

Substitute back

u = 1 + e^{x}

= - (1 + e^{x} - ln ∣ 1 + e^{x} ∣)

Simplify

= - 1 - e^{x} + ln ∣ 1 + e^{x} ∣

Add a constant to the solution

= - 1 - e^{x} + ln ∣ 1 + e^{x} ∣ + C

x \to 0 + lim ((tan (6 x))^{x})

\frac{d y}{d x} = x (x - 1)

\frac{d}{d x} (2 - \frac{1}{x})

\int (\frac{1}{ln ( x )} + ln (ln (x))) d x

h \to 2 + lim (e^{\frac{2}{h}})

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