What is the integral of (e^{4x}+e^{2x}+1)/(e^x-1) ?

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

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integral of (e^{4x}+e^{2x}+1)/(e^x-1)

\int \frac{e ^{4 x} + e ^{2 x} + 1}{e ^{x} - 1} d x

- x + 4 e^{x} + 3 ln ∣ e^{x} - 1 ∣ + \frac{1}{3} (e^{x} - 1)^{3} + \frac{3}{2} (e^{x} - 1)^{2} - 4 + C

Solution steps

\int \frac{e ^{4 x} + e ^{2 x} + 1}{e ^{x} - 1} d x

Expand

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

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{e ^{4 x}}{e ^{x} - 1} d x + \int \frac{e ^{2 x}}{e ^{x} - 1} d x + \int \frac{1}{e ^{x} - 1} d x

\int \frac{e ^{4 x}}{e ^{x} - 1} d x = \frac{( e ^{x} - 1 ) ^{3}}{3} + \frac{3 ( e ^{x} - 1 ) ^{2}}{2} + 3 (e^{x} - 1) + ln ∣ e^{x} - 1 ∣

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

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

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

Simplify

\frac{( e ^{x} - 1 ) ^{3}}{3} + \frac{3 ( e ^{x} - 1 ) ^{2}}{2} + 3 (e^{x} - 1) + ln ∣ e^{x} - 1 ∣ + e^{x} - 1 + ln ∣ e^{x} - 1 ∣ + ln ∣ e^{x} - 1 ∣ - x : - x + 4 e^{x} + 3 ln ∣ e^{x} - 1 ∣ + \frac{1}{3} (e^{x} - 1)^{3} + \frac{3}{2} (e^{x} - 1)^{2} - 4

= - x + 4 e^{x} + 3 ln ∣ e^{x} - 1 ∣ + \frac{1}{3} (e^{x} - 1)^{3} + \frac{3}{2} (e^{x} - 1)^{2} - 4

Add a constant to the solution

= - x + 4 e^{x} + 3 ln ∣ e^{x} - 1 ∣ + \frac{1}{3} (e^{x} - 1)^{3} + \frac{3}{2} (e^{x} - 1)^{2} - 4 + C

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