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

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

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

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

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

Solution steps

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

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

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

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 t}}{e ^{2 t} + 1} d t + \int 1 d t

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

\int 1 d t = t

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

Add a constant to the solution

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

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