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

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

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

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

\frac{1}{5} ln ∣ e^{x} - 1 ∣ - \frac{1}{5} ln ∣ e^{x} + 4 ∣ + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{( u - 1 ) ( u + 4 )} d u

Take the partial fraction of

\frac{1}{( u - 1 ) ( u + 4 )} : \frac{1}{5 ( u - 1 )} - \frac{1}{5 ( u + 4 )}

= \int \frac{1}{5 ( u - 1 )} - \frac{1}{5 ( u + 4 )} 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}{5 ( u - 1 )} d u - \int \frac{1}{5 ( u + 4 )} d u

\int \frac{1}{5 ( u - 1 )} d u = \frac{1}{5} ln ∣ u - 1 ∣

\int \frac{1}{5 ( u + 4 )} d u = \frac{1}{5} ln ∣ u + 4 ∣

= \frac{1}{5} ln ∣ u - 1 ∣ - \frac{1}{5} ln ∣ u + 4 ∣

Substitute back

u = e^{x}

= \frac{1}{5} ln ∣ e^{x} - 1 ∣ - \frac{1}{5} ln ∣ e^{x} + 4 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ e^{x} - 1 ∣ - \frac{1}{5} ln ∣ e^{x} + 4 ∣ + C

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