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

The integral of (e^x)/(e^{2x)+4e^x+4} is -1/(e^x+2)+C

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

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

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

Solution steps

\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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

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

Add a constant to the solution

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

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

\int_{6}^{7} (t t - 6) d t

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

y^{^{'}} = 3 (y + 1)

\frac{cos ( x )}{sin ( x )}

What is the integral of (e^x)/(e^{2x)+4e^x+4} ?
The integral of (e^x)/(e^{2x)+4e^x+4} is -1/(e^x+2)+C