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

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

Solution steps

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

Expand

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

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

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

\int \frac{1}{e ^{2 x} - 3} d x = \frac{1}{6} (ln e^{2 x} - 3 - 2 x)

= \frac{1}{2} ln e^{2 x} - 3 + \frac{1}{6} (ln e^{2 x} - 3 - 2 x)

Simplify

\frac{1}{2} ln e^{2 x} - 3 + \frac{1}{6} (ln e^{2 x} - 3 - 2 x) : \frac{2}{3} ln e^{2 x} - 3 - \frac{1}{3} x

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

Add a constant to the solution

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

\frac{1}{x - 1}, 25

t \to 1 lim (sin (- 5 π t))

\int_{1}^{5} \frac{x}{5 - x} d x

\frac{9}{4} t^{2} e^{2 t} e^{3 t}

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