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

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

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

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

ln e^{2 x} + 1 - x + C

Solution steps

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

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

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

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

\int 1 d x = x

= ln e^{2 x} + 1 - x

Add a constant to the solution

= ln e^{2 x} + 1 - x + C

x \to 6 + lim (\frac{8 + x}{6 - x})

\int x^{3} e^{(x^{2})} d x

\int (8 x + 3) d x

\frac{\partial}{\partial y} (e^{x y} ln (y))

\frac{3}{1 - 6 x}

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