What is the integral of (e^x)/(e^{2x)+4e^x+3} ?

The integral of (e^x)/(e^{2x)+4e^x+3} is 1/2 ln|e^x+1|-1/2 ln|e^x+3|+C

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integral of (e^x)/(e^{2x)+4e^x+3}

\int \frac{e ^{x}}{e ^{2 x} + 4 e ^{x} + 3} d x

\frac{1}{2} ln ∣ e^{x} + 1 ∣ - \frac{1}{2} ln ∣ e^{x} + 3 ∣ + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} + 4 e ^{x} + 3} d x

Apply u-substitution

= \int \frac{1}{u ^{2} + 4 u + 3} d u

Take the partial fraction of

\frac{1}{u ^{2} + 4 u + 3} : \frac{1}{2 ( u + 1 )} - \frac{1}{2 ( u + 3 )}

= \int \frac{1}{2 ( u + 1 )} - \frac{1}{2 ( u + 3 )} 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}{2 ( u + 1 )} d u - \int \frac{1}{2 ( u + 3 )} d u

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

\int \frac{1}{2 ( u + 3 )} d u = \frac{1}{2} ln ∣ u + 3 ∣

= \frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u + 3 ∣

Substitute back

u = e^{x}

= \frac{1}{2} ln ∣ e^{x} + 1 ∣ - \frac{1}{2} ln ∣ e^{x} + 3 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ e^{x} + 1 ∣ - \frac{1}{2} ln ∣ e^{x} + 3 ∣ + C

What is the integral of (e^x)/(e^{2x)+4e^x+3} ?
The integral of (e^x)/(e^{2x)+4e^x+3} is 1/2 ln|e^x+1|-1/2 ln|e^x+3|+C