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

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

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

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

e^{x} + x + C

Solution steps

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

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

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

Simplify

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

= \int 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 e^{x} d x + \int 1 d x

\int e^{x} d x = e^{x}

\int 1 d x = x

= e^{x} + x

Add a constant to the solution

= e^{x} + x + C

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