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

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

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

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

- e^{x} + x + C

Solution steps

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

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

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

Simplify

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

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