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

The integral of (e^x)/(e^{3x)-e^{2x}+e^x-1} is 1/2 ln|e^x-1|-1/2 arctan(e^x)-1/4 ln|e^{2x}+1|+C

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integral of (e^x)/(e^{3x)-e^{2x}+e^x-1}

\int \frac{e ^{x}}{e ^{3 x} - e ^{2 x} + e ^{x} - 1} d x

\frac{1}{2} ln ∣ e^{x} - 1 ∣ - \frac{1}{2} arctan (e^{x}) - \frac{1}{4} ln e^{2 x} + 1 + C

Solution steps

\int \frac{e ^{x}}{e ^{3 x} - e ^{2 x} + e ^{x} - 1} d x

Take the partial fraction of

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

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

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

\int \frac{1}{2 ( e ^{x} - 1 )} d x = \frac{1}{2} (ln ∣ e^{x} - 1 ∣ - x)

\int \frac{- e ^{x} + 1}{2 ( e ^{2 x} + 1 )} d x = \frac{1}{2} (- arctan (e^{x}) - \frac{1}{2} ln e^{2 x} + 1 + x)

= \frac{1}{2} (ln ∣ e^{x} - 1 ∣ - x) + \frac{1}{2} (- arctan (e^{x}) - \frac{1}{2} ln e^{2 x} + 1 + x)

Simplify

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

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

Add a constant to the solution

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

What is the integral of (e^x)/(e^{3x)-e^{2x}+e^x-1} ?
The integral of (e^x)/(e^{3x)-e^{2x}+e^x-1} is 1/2 ln|e^x-1|-1/2 arctan(e^x)-1/4 ln|e^{2x}+1|+C