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

The integral of 1/(e^{3t)+e^t} is -1/(e^t)-arctan(e^t)+C

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integral of 1/(e^{3t)+e^t}

\int \frac{1}{e ^{3 t} + e ^{t}} d t

- \frac{1}{e ^{t}} - arctan (e^{t}) + C

Solution steps

\int \frac{1}{e ^{3 t} + e ^{t}} d t

Factor

\frac{1}{e ^{3 t} + e ^{t}}

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

Take the partial fraction of

\frac{1}{e ^{t} ( e ^{2 t} + 1 )} : \frac{1}{e ^{t}} - \frac{e ^{t}}{e ^{2 t} + 1}

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

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

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

= - \frac{1}{e ^{t}} - arctan (e^{t})

Add a constant to the solution

= - \frac{1}{e ^{t}} - arctan (e^{t}) + C

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What is the integral of 1/(e^{3t)+e^t} ?
The integral of 1/(e^{3t)+e^t} is -1/(e^t)-arctan(e^t)+C