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

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

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

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

- \frac{4}{e ^{x} + 2} + C

Solution steps

\int \frac{4 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

= 4 \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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= 4 (- \frac{1}{v})

Substitute back

= 4 (- \frac{1}{e ^{x} + 2})

Simplify

4 (- \frac{1}{e ^{x} + 2}) : - \frac{4}{e ^{x} + 2}

= - \frac{4}{e ^{x} + 2}

Add a constant to the solution

= - \frac{4}{e ^{x} + 2} + C

\frac{d y}{d t} - \frac{1}{2} y = t y^{- 1}

\int e^{- 0.12 x} d x

\int \frac{1}{z ^{2}} d z

f (x) = - 5 e^{- 5 x + 2}

n = 2 \sum \infty \frac{1}{n ^{\frac{1}{2}}}

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