What is the integral of (e^x)/(e^{2x)+12e^x+27} ?

The integral of (e^x)/(e^{2x)+12e^x+27} is 1/6 ln|e^x+3|-1/6 ln|e^x+9|+C

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integral of (e^x)/(e^{2x)+12e^x+27}

\int \frac{e ^{x}}{e ^{2 x} + 12 e ^{x} + 27} d x

\frac{1}{6} ln ∣ e^{x} + 3 ∣ - \frac{1}{6} ln ∣ e^{x} + 9 ∣ + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} + 12 e ^{x} + 27} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 12 u + 27} d u

= \int \frac{1}{u ^{2} + 12 u + 27} d u

Take the partial fraction of

\frac{1}{u ^{2} + 12 u + 27} : \frac{1}{6 ( u + 3 )} - \frac{1}{6 ( u + 9 )}

= \int \frac{1}{6 ( u + 3 )} - \frac{1}{6 ( u + 9 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{6 ( u + 3 )} d u - \int \frac{1}{6 ( u + 9 )} d u

\int \frac{1}{6 ( u + 3 )} d u = \frac{1}{6} ln ∣ u + 3 ∣

\int \frac{1}{6 ( u + 9 )} d u = \frac{1}{6} ln ∣ u + 9 ∣

= \frac{1}{6} ln ∣ u + 3 ∣ - \frac{1}{6} ln ∣ u + 9 ∣

Substitute back

u = e^{x}

= \frac{1}{6} ln ∣ e^{x} + 3 ∣ - \frac{1}{6} ln ∣ e^{x} + 9 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ e^{x} + 3 ∣ - \frac{1}{6} ln ∣ e^{x} + 9 ∣ + C

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What is the integral of (e^x)/(e^{2x)+12e^x+27} ?
The integral of (e^x)/(e^{2x)+12e^x+27} is 1/6 ln|e^x+3|-1/6 ln|e^x+9|+C