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

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

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

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

- \frac{2}{e ^{x} + 1} + C

Solution steps

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

Take the constant out:

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

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

Factor

\frac{e ^{x}}{e ^{2 x} + 2 e ^{x} + 1}

= 2 \cdot \int \frac{e ^{x}}{( e ^{x} + 1 ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= 2 (- \frac{1}{u})

Substitute back

u = e^{x} + 1

= 2 (- \frac{1}{e ^{x} + 1})

Simplify

2 (- \frac{1}{e ^{x} + 1}) : - \frac{2}{e ^{x} + 1}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{2 x - 1}{3 x ^{2}})

x y^{^{'}} = y + x e^{\frac{y}{x}}

e^{3 x}, e^{x}, [0, 1]

\int_{- 6}^{8} 10 d x

\int_{- 4}^{2} - x^{2} - 2 x + 8 d x

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