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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{e ^{x} - 1} d x = ln ∣ e^{x} - 1 ∣ - x

\int \frac{2}{e ^{x} - 2} d x = - x + ln ∣ e^{x} - 2 ∣

= - (ln ∣ e^{x} - 1 ∣ - x) - x + ln ∣ e^{x} - 2 ∣

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (tan^{2} (3 x) + 1)

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

n = 7 \sum \infty \frac{1}{6 n}

y = x^{2} + 5 x, x = 8

\int (x^{2} - 2 x + 3) d x

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