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

The integral of 1/(e^{2x)+e^x-2} is -1/2 x+1/3 ln|e^x-1|+1/6 ln|e^x+2|+C

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

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

- \frac{1}{2} x + \frac{1}{3} ln ∣ e^{x} - 1 ∣ + \frac{1}{6} ln ∣ e^{x} + 2 ∣ + C

Solution steps

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

Apply u-substitution

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

Take the partial fraction of

\frac{1}{u ( u ^{2} + u - 2 )} : - \frac{1}{2 u} + \frac{1}{3 ( u - 1 )} + \frac{1}{6 ( u + 2 )}

= \int - \frac{1}{2 u} + \frac{1}{3 ( u - 1 )} + \frac{1}{6 ( 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}{2 u} d u + \int \frac{1}{3 ( u - 1 )} d u + \int \frac{1}{6 ( u + 2 )} d u

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

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

\int \frac{1}{6 ( u + 2 )} d u = \frac{1}{6} ln ∣ u + 2 ∣

= - \frac{1}{2} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 1 ∣ + \frac{1}{6} ln ∣ u + 2 ∣

Substitute back

u = e^{x}

= - \frac{1}{2} ln ∣ e^{x} ∣ + \frac{1}{3} ln ∣ e^{x} - 1 ∣ + \frac{1}{6} ln ∣ e^{x} + 2 ∣

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

= - \frac{1}{2} x + \frac{1}{3} ln ∣ e^{x} - 1 ∣ + \frac{1}{6} ln ∣ e^{x} + 2 ∣

Add a constant to the solution

= - \frac{1}{2} x + \frac{1}{3} ln ∣ e^{x} - 1 ∣ + \frac{1}{6} ln ∣ e^{x} + 2 ∣ + C

What is the integral of 1/(e^{2x)+e^x-2} ?
The integral of 1/(e^{2x)+e^x-2} is -1/2 x+1/3 ln|e^x-1|+1/6 ln|e^x+2|+C