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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the partial fraction of

\frac{1}{u + 2 u ^{2} - 1} : \frac{2}{3 ( 2 u - 1 )} - \frac{1}{3 ( u + 1 )}

= \int \frac{2}{3 ( 2 u - 1 )} - \frac{1}{3 ( u + 1 )} d u

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{2}{3 ( 2 u - 1 )} d u - \int \frac{1}{3 ( u + 1 )} d u

\int \frac{2}{3 ( 2 u - 1 )} d u = \frac{1}{3} ln ∣ 2 u - 1 ∣

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

= \frac{1}{3} ln ∣ 2 u - 1 ∣ - \frac{1}{3} ln ∣ u + 1 ∣

Substitute back

u = e^{x}

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

Add a constant to the solution

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

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