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

The integral of 1/(e^{2y)-2e^y} is -1/4 y+1/(2e^y)+1/4 ln|e^y-2|+C

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

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

- \frac{1}{4} y + \frac{1}{2 e ^{y}} + \frac{1}{4} ln ∣ e^{y} - 2 ∣ + C

Solution steps

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

Factor

\frac{1}{e ^{2 y} - 2 e ^{y}}

= \int \frac{1}{e ^{y} ( e ^{y} - 2 )} d y

Apply u-substitution

: \int \frac{1}{u ^{2} ( u - 2 )} d u

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

Take the partial fraction of

\frac{1}{u ^{2} ( u - 2 )} : - \frac{1}{4 u} - \frac{1}{2 u ^{2}} + \frac{1}{4 ( u - 2 )}

= \int - \frac{1}{4 u} - \frac{1}{2 u ^{2}} + \frac{1}{4 ( u - 2 )} 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{1}{4 u} d u - \int \frac{1}{2 u ^{2}} d u + \int \frac{1}{4 ( u - 2 )} d u

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

\int \frac{1}{4 ( u - 2 )} d u = \frac{1}{4} ln ∣ u - 2 ∣

= - \frac{1}{4} ln ∣ u ∣ - (- \frac{1}{2 u}) + \frac{1}{4} ln ∣ u - 2 ∣

Substitute back

u = e^{y}

= - \frac{1}{4} ln ∣ e^{y} ∣ - (- \frac{1}{2 e ^{y}}) + \frac{1}{4} ln ∣ e^{y} - 2 ∣

Simplify

- \frac{1}{4} ln ∣ e^{y} ∣ - (- \frac{1}{2 e ^{y}}) + \frac{1}{4} ln ∣ e^{y} - 2 ∣ : - \frac{1}{4} y + \frac{1}{2 e ^{y}} + \frac{1}{4} ln ∣ e^{y} - 2 ∣

= - \frac{1}{4} y + \frac{1}{2 e ^{y}} + \frac{1}{4} ln ∣ e^{y} - 2 ∣

Add a constant to the solution

= - \frac{1}{4} y + \frac{1}{2 e ^{y}} + \frac{1}{4} ln ∣ e^{y} - 2 ∣ + C

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What is the integral of 1/(e^{2y)-2e^y} ?
The integral of 1/(e^{2y)-2e^y} is -1/4 y+1/(2e^y)+1/4 ln|e^y-2|+C