What is the integral of (1/(e^x+e^{2x)})/(-e^{-3x)} ?

The integral of (1/(e^x+e^{2x)})/(-e^{-3x)} is -1-e^x+ln|1+e^x|+C

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integral of (1/(e^x+e^{2x)})/(-e^{-3x)}

\int \frac{\frac{1}{e ^{x} + e ^{2 x}}}{- e ^{- 3 x}} d x

- 1 - e^{x} + ln ∣ 1 + e^{x} ∣ + C

Solution steps

\int \frac{\frac{1}{e ^{x} + e ^{2 x}}}{- e ^{- 3 x}} d x

Take the constant out:

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

= - \int \frac{\frac{1}{e ^{x} + e ^{2 x}}}{e ^{- 3 x}} d x

\frac{\frac{1}{e ^{x} + e ^{2 x}}}{e ^{- 3 x}} = \frac{1}{e ^{x} + e ^{2 x}} \cdot \frac{1}{e ^{- 3 x}}

= - \int \frac{1}{e ^{x} + e ^{2 x}} \cdot \frac{1}{e ^{- 3 x}} d x

Simplify

\frac{1}{e ^{x} + e ^{2 x}} \cdot \frac{1}{e ^{- 3 x}} : \frac{e ^{2 x}}{1 + e ^{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 2 - lim (\frac{x - 1}{∣ x - 1 ∣})

\frac{d}{d x} (3 x^{3} y^{2})

\frac{d y}{d x} = \frac{2 x + sec ^{2} ( x )}{4 y}

\frac{\partial}{\partial y} (y ln (x))

\frac{d}{d x} (40 x^{4} + 180 x^{3} + 240 x^{2} + 100 x)

What is the integral of (1/(e^x+e^{2x)})/(-e^{-3x)} ?
The integral of (1/(e^x+e^{2x)})/(-e^{-3x)} is -1-e^x+ln|1+e^x|+C