What is the integral of (e^x-1)/(e^{2x)-5e^x+6} ?

The integral of (e^x-1)/(e^{2x)-5e^x+6} is -1/6 x-1/2 ln|e^x-2|+2/3 ln|e^x-3|+C

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integral of (e^x-1)/(e^{2x)-5e^x+6}

\int \frac{e ^{x} - 1}{e ^{2 x} - 5 e ^{x} + 6} d x

- \frac{1}{6} x - \frac{1}{2} ln ∣ e^{x} - 2 ∣ + \frac{2}{3} ln ∣ e^{x} - 3 ∣ + C

Solution steps

\int \frac{e ^{x} - 1}{e ^{2 x} - 5 e ^{x} + 6} d x

Expand

\frac{e ^{x} - 1}{e ^{2 x} - 5 e ^{x} + 6} : \frac{e ^{x}}{e ^{2 x} - 5 e ^{x} + 6} - \frac{1}{e ^{2 x} - 5 e ^{x} + 6}

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{e ^{x}}{e ^{2 x} - 5 e ^{x} + 6} d x - \int \frac{1}{e ^{2 x} - 5 e ^{x} + 6} d x

\int \frac{e ^{x}}{e ^{2 x} - 5 e ^{x} + 6} d x = - ln ∣ e^{x} - 2 ∣ + ln ∣ e^{x} - 3 ∣

\int \frac{1}{e ^{2 x} - 5 e ^{x} + 6} d x = \frac{1}{6} x - \frac{1}{2} ln ∣ e^{x} - 2 ∣ + \frac{1}{3} ln ∣ e^{x} - 3 ∣

= - ln ∣ e^{x} - 2 ∣ + ln ∣ e^{x} - 3 ∣ - (\frac{1}{6} x - \frac{1}{2} ln ∣ e^{x} - 2 ∣ + \frac{1}{3} ln ∣ e^{x} - 3 ∣)

Simplify

- ln ∣ e^{x} - 2 ∣ + ln ∣ e^{x} - 3 ∣ - (\frac{1}{6} x - \frac{1}{2} ln ∣ e^{x} - 2 ∣ + \frac{1}{3} ln ∣ e^{x} - 3 ∣) : - \frac{1}{6} x - \frac{1}{2} ln ∣ e^{x} - 2 ∣ + \frac{2}{3} ln ∣ e^{x} - 3 ∣

= - \frac{1}{6} x - \frac{1}{2} ln ∣ e^{x} - 2 ∣ + \frac{2}{3} ln ∣ e^{x} - 3 ∣

Add a constant to the solution

= - \frac{1}{6} x - \frac{1}{2} ln ∣ e^{x} - 2 ∣ + \frac{2}{3} ln ∣ e^{x} - 3 ∣ + C

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What is the integral of (e^x-1)/(e^{2x)-5e^x+6} ?
The integral of (e^x-1)/(e^{2x)-5e^x+6} is -1/6 x-1/2 ln|e^x-2|+2/3 ln|e^x-3|+C