What is the integral of (e^{2x}+1)/(e^{2x)-9} ?

The integral of (e^{2x}+1)/(e^{2x)-9} is 5/9 ln|e^{2x}-9|-1/9 x+C

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integral of (e^{2x}+1)/(e^{2x)-9}

\int \frac{e ^{2 x} + 1}{e ^{2 x} - 9} d x

\frac{5}{9} ln e^{2 x} - 9 - \frac{1}{9} x + C

Solution steps

\int \frac{e ^{2 x} + 1}{e ^{2 x} - 9} d x

Expand

\frac{e ^{2 x} + 1}{e ^{2 x} - 9} : \frac{e ^{2 x}}{e ^{2 x} - 9} + \frac{1}{e ^{2 x} - 9}

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 ^{2 x}}{e ^{2 x} - 9} d x + \int \frac{1}{e ^{2 x} - 9} d x

\int \frac{e ^{2 x}}{e ^{2 x} - 9} d x = \frac{1}{2} ln e^{2 x} - 9

\int \frac{1}{e ^{2 x} - 9} d x = \frac{1}{18} (ln e^{2 x} - 9 - 2 x)

= \frac{1}{2} ln e^{2 x} - 9 + \frac{1}{18} (ln e^{2 x} - 9 - 2 x)

Simplify

\frac{1}{2} ln e^{2 x} - 9 + \frac{1}{18} (ln e^{2 x} - 9 - 2 x) : \frac{5}{9} ln e^{2 x} - 9 - \frac{1}{9} x

= \frac{5}{9} ln e^{2 x} - 9 - \frac{1}{9} x

Add a constant to the solution

= \frac{5}{9} ln e^{2 x} - 9 - \frac{1}{9} x + C

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What is the integral of (e^{2x}+1)/(e^{2x)-9} ?
The integral of (e^{2x}+1)/(e^{2x)-9} is 5/9 ln|e^{2x}-9|-1/9 x+C