What is the integral of (e^{2x}-1)/(e^{2x)+3} ?

The integral of (e^{2x}-1)/(e^{2x)+3} is -1/3 x+2/3 ln|e^{2x}+3|+C

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integral of (e^{2x}-1)/(e^{2x)+3}

\int \frac{e ^{2 x} - 1}{e ^{2 x} + 3} d x

- \frac{1}{3} x + \frac{2}{3} ln e^{2 x} + 3 + C

Solution steps

\int \frac{e ^{2 x} - 1}{e ^{2 x} + 3} d x

\frac{e ^{2 x} - 1}{e ^{2 x} + 3} = - \frac{4}{e ^{2 x} + 3} + 1

= \int - \frac{4}{e ^{2 x} + 3} + 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{4}{e ^{2 x} + 3} d x + \int 1 d x

\int \frac{4}{e ^{2 x} + 3} d x = 2 (\frac{2}{3} x - \frac{1}{3} ln e^{2 x} + 3)

\int 1 d x = x

= - 2 (\frac{2}{3} x - \frac{1}{3} ln e^{2 x} + 3) + x

Simplify

- 2 (\frac{2}{3} x - \frac{1}{3} ln e^{2 x} + 3) + x : - \frac{1}{3} x + \frac{2}{3} ln e^{2 x} + 3

= - \frac{1}{3} x + \frac{2}{3} ln e^{2 x} + 3

Add a constant to the solution

= - \frac{1}{3} x + \frac{2}{3} ln e^{2 x} + 3 + C

x \to 3 lim (\frac{x + 1}{x ^{2} - 9})

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

\frac{d}{d x} (- \frac{1}{64} sin (\frac{x}{8}))

\frac{d}{d t} (a cos (wt) + b sin (wt))

\int (e^{5 x}) (cos (- 2 x)) d x

What is the integral of (e^{2x}-1)/(e^{2x)+3} ?
The integral of (e^{2x}-1)/(e^{2x)+3} is -1/3 x+2/3 ln|e^{2x}+3|+C