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

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

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

\int \frac{e ^{x}}{e ^{2 x} - 2 e ^{x} - 3} d x

- \frac{1}{4} ln ∣ e^{x} + 1 ∣ + \frac{1}{4} ln ∣ e^{x} - 3 ∣ + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} - 2 e ^{x} - 3} d x

Take the partial fraction of

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

= \int \frac{1}{4 ( e ^{x} + 1 )} + \frac{3}{4 ( e ^{x} - 3 )} d x

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}{4 ( e ^{x} + 1 )} d x + \int \frac{3}{4 ( e ^{x} - 3 )} d x

\int \frac{1}{4 ( e ^{x} + 1 )} d x = \frac{1}{4} (- ln ∣ e^{x} + 1 ∣ + x)

\int \frac{3}{4 ( e ^{x} - 3 )} d x = \frac{1}{4} (- x + ln ∣ e^{x} - 3 ∣)

= \frac{1}{4} (- ln ∣ e^{x} + 1 ∣ + x) + \frac{1}{4} (- x + ln ∣ e^{x} - 3 ∣)

Simplify

\frac{1}{4} (- ln ∣ e^{x} + 1 ∣ + x) + \frac{1}{4} (- x + ln ∣ e^{x} - 3 ∣) : - \frac{1}{4} ln ∣ e^{x} + 1 ∣ + \frac{1}{4} ln ∣ e^{x} - 3 ∣

= - \frac{1}{4} ln ∣ e^{x} + 1 ∣ + \frac{1}{4} ln ∣ e^{x} - 3 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ e^{x} + 1 ∣ + \frac{1}{4} ln ∣ e^{x} - 3 ∣ + C

\frac{x ^{3} - x}{1 + x ^{2}}

f (x) = x^{2} - x, (- 4, 20)

x \cdot y \cdot \frac{d y}{d x} - y^{2} = 2 \cdot x^{2}

n = 4 \sum \infty \frac{1}{n ( n + 1 )}

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

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