What is the integral of (e^x-1)/(e^{2x)+4} ?

The integral of (e^x-1)/(e^{2x)+4} is 1/2 arctan((e^x)/2)-1/8 (-ln|e^{2x}+4|+2x)+C

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integral of (e^x-1)/(e^{2x)+4}

\int \frac{e ^{x} - 1}{e ^{2 x} + 4} d x

\frac{1}{2} arctan (\frac{e ^{x}}{2}) - \frac{1}{8} (- ln e^{2 x} + 4 + 2 x) + C

Solution steps

\int \frac{e ^{x} - 1}{e ^{2 x} + 4} d x

Expand

\frac{e ^{x} - 1}{e ^{2 x} + 4} : \frac{e ^{x}}{e ^{2 x} + 4} - \frac{1}{e ^{2 x} + 4}

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 ^{x}}{e ^{2 x} + 4} d x - \int \frac{1}{e ^{2 x} + 4} d x

\int \frac{e ^{x}}{e ^{2 x} + 4} d x = \frac{1}{2} arctan (\frac{e ^{x}}{2})

\int \frac{1}{e ^{2 x} + 4} d x = \frac{1}{8} (- ln e^{2 x} + 4 + 2 x)

= \frac{1}{2} arctan (\frac{e ^{x}}{2}) - \frac{1}{8} (- ln e^{2 x} + 4 + 2 x)

Add a constant to the solution

= \frac{1}{2} arctan (\frac{e ^{x}}{2}) - \frac{1}{8} (- ln e^{2 x} + 4 + 2 x) + C

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What is the integral of (e^x-1)/(e^{2x)+4} ?
The integral of (e^x-1)/(e^{2x)+4} is 1/2 arctan((e^x)/2)-1/8 (-ln|e^{2x}+4|+2x)+C