What is the integral of (e^{2x}+e^{4x})/(e^x-e^{5x)} ?

The integral of (e^{2x}+e^{4x})/(e^x-e^{5x)} is 1/2 ln|e^x+1|-1/2 ln|e^x-1|+C

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integral of (e^{2x}+e^{4x})/(e^x-e^{5x)}

\int \frac{e ^{2 x} + e ^{4 x}}{e ^{x} - e ^{5 x}} d x

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

Solution steps

\int \frac{e ^{2 x} + e ^{4 x}}{e ^{x} - e ^{5 x}} d x

Take the partial fraction of

\frac{e ^{2 x} + e ^{4 x}}{e ^{x} - e ^{5 x}} : - \frac{1}{2 ( e ^{x} + 1 )} - \frac{1}{2 ( e ^{x} - 1 )}

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

(9 y^{2} - x y) d x + (x^{2}) d y = 0

\int_{1}^{7} \frac{1}{x - 2} d x

\int \frac{x ^{3}}{x ^{2} + 9} d x

4^{- 3} (\frac{2}{s ^{7}} + 3 s)^{4}

n = 0 \sum \infty \frac{1}{2} (\frac{1}{2})^{n}

What is the integral of (e^{2x}+e^{4x})/(e^x-e^{5x)} ?
The integral of (e^{2x}+e^{4x})/(e^x-e^{5x)} is 1/2 ln|e^x+1|-1/2 ln|e^x-1|+C