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

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

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

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

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

Solution steps

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

Long division

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

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

\int 1 d x = x

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

= x - \frac{2}{e ^{x} - 1}

Add a constant to the solution

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

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