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

The integral of (e^x)/(e^{3x)-2e^{2x}+e^x-2} is 1/5 ln|e^x-2|-2/5 arctan(e^x)-1/10 ln|e^{2x}+1|+C

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

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

\frac{1}{5} ln ∣ e^{x} - 2 ∣ - \frac{2}{5} arctan (e^{x}) - \frac{1}{10} ln e^{2 x} + 1 + C

Solution steps

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

Take the partial fraction of

\frac{e ^{x}}{e ^{3 x} - 2 e ^{2 x} + e ^{x} - 2} : \frac{2}{5 ( e ^{x} - 2 )} + \frac{- 2 e ^{x} + 1}{5 ( e ^{2 x} + 1 )}

= \int \frac{2}{5 ( e ^{x} - 2 )} + \frac{- 2 e ^{x} + 1}{5 ( 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{2}{5 ( e ^{x} - 2 )} d x + \int \frac{- 2 e ^{x} + 1}{5 ( e ^{2 x} + 1 )} d x

\int \frac{2}{5 ( e ^{x} - 2 )} d x = - \frac{1}{5} (x - ln ∣ e^{x} - 2 ∣)

\int \frac{- 2 e ^{x} + 1}{5 ( e ^{2 x} + 1 )} d x = \frac{1}{5} (- 2 arctan (e^{x}) - \frac{1}{2} ln e^{2 x} + 1 + x)

= - \frac{1}{5} (x - ln ∣ e^{x} - 2 ∣) + \frac{1}{5} (- 2 arctan (e^{x}) - \frac{1}{2} ln e^{2 x} + 1 + x)

Simplify

- \frac{1}{5} (x - ln ∣ e^{x} - 2 ∣) + \frac{1}{5} (- 2 arctan (e^{x}) - \frac{1}{2} ln e^{2 x} + 1 + x) : \frac{1}{5} ln ∣ e^{x} - 2 ∣ - \frac{2}{5} arctan (e^{x}) - \frac{1}{10} ln e^{2 x} + 1

= \frac{1}{5} ln ∣ e^{x} - 2 ∣ - \frac{2}{5} arctan (e^{x}) - \frac{1}{10} ln e^{2 x} + 1

Add a constant to the solution

= \frac{1}{5} ln ∣ e^{x} - 2 ∣ - \frac{2}{5} arctan (e^{x}) - \frac{1}{10} ln e^{2 x} + 1 + C

\frac{\partial}{\partial y} (6 x y^{2})

\int 4 x ln (7 x) d x

\frac{\partial}{\partial x} (ln (2 x + y + 1))

y^{^{'}} + 3 y = 8 x, y (0) = 4

f (x) = 1 - 9 x^{2}, at x = 3

What is the integral of (e^x)/(e^{3x)-2e^{2x}+e^x-2} ?
The integral of (e^x)/(e^{3x)-2e^{2x}+e^x-2} is 1/5 ln|e^x-2|-2/5 arctan(e^x)-1/10 ln|e^{2x}+1|+C