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

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

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

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

- 2 arctan (e^{x}) + e^{x} + C

Solution steps

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

Expand

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

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

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

= - arctan (e^{x}) + e^{x} - arctan (e^{x})

Simplify

= - 2 arctan (e^{x}) + e^{x}

Add a constant to the solution

= - 2 arctan (e^{x}) + e^{x} + C

x \to - 6 - lim (tan (\frac{π x}{4}))

\int e^{s i n (x)} sin (x) cos (x) d x

\int 6 x^{4} d x

y^{^{'}} (x) = (\frac{9 x}{9 y ^{2} - 7})^{2}, y (1) = 0

\frac{18}{( 3 x + 1 ) ^{3}}

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