What is the integral of (2e^x)/(e^{2x)+4e^x+4} ?

The integral of (2e^x)/(e^{2x)+4e^x+4} is -2/(e^x+2)+C

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integral of (2e^x)/(e^{2x)+4e^x+4}

\int \frac{2 e ^{x}}{e ^{2 x} + 4 e ^{x} + 4} d x

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

Solution steps

\int \frac{2 e ^{x}}{e ^{2 x} + 4 e ^{x} + 4} 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} + 4 e ^{x} + 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 4 u + 4} d u

= 2 \cdot \int \frac{1}{u ^{2} + 4 u + 4} d u

Complete the square

u^{2} + 4 u + 4 : (u + 2)^{2}

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

x \to \infty lim (\frac{7}{x} - 2)

\int x^{2} + 5 e^{x} + C d x

\int_{0}^{1} (10 + x x) d x

\frac{4 s e ^{- 4 s}}{3 s ^{2} + 27}

What is the integral of (2e^x)/(e^{2x)+4e^x+4} ?
The integral of (2e^x)/(e^{2x)+4e^x+4} is -2/(e^x+2)+C