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

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

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

\int \frac{e ^{2 t}}{e ^{2 t} - 1} d t

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

Solution steps

\int \frac{e ^{2 t}}{e ^{2 t} - 1} d t

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = e^{2 t} - 1

= \frac{1}{2} ln e^{2 t} - 1

Add a constant to the solution

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

\frac{d}{d x} (ln (\frac{1 + tan ( x )}{1 + sin ( x )}))

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{19}{2}}} d t

f (x) = x^{3} sin (x)

x \to \infty lim (x b)

\int \frac{3 x ^{3}}{7 - 9 x ^{4}} d x

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