What is the integral of (e^x)/(e^{2x)+8e^x+16} ?

The integral of (e^x)/(e^{2x)+8e^x+16} is -1/(e^x+4)+C

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integral of (e^x)/(e^{2x)+8e^x+16}

\int \frac{e ^{x}}{e ^{2 x} + 8 e ^{x} + 16} d x

- \frac{1}{e ^{x} + 4} + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} + 8 e ^{x} + 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 8 u + 16} d u

= \int \frac{1}{u ^{2} + 8 u + 16} d u

Take the partial fraction of

\frac{1}{u ^{2} + 8 u + 16} : \frac{1}{( u + 4 ) ^{2}}

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

Add a constant to the solution

= - \frac{1}{e ^{x} + 4} + C

\int ln (x^{2} + 4) d x

\int 6 x + 4 d x

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

n = 4 \sum \infty \frac{1}{n ^{2} - 5 n + 6}

x \to - 4 lim (- \frac{4}{x + 4})

What is the integral of (e^x)/(e^{2x)+8e^x+16} ?
The integral of (e^x)/(e^{2x)+8e^x+16} is -1/(e^x+4)+C