What is the integral of (e^x)/((e^x-6)(e^x+1)) ?

The integral of (e^x)/((e^x-6)(e^x+1)) is -1/7 ln|e^x+1|+1/7 ln|e^x-6|+C

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integral of (e^x)/((e^x-6)(e^x+1))

\int \frac{e ^{x}}{( e ^{x} - 6 ) ( e ^{x} + 1 )} d x

- \frac{1}{7} ln ∣ e^{x} + 1 ∣ + \frac{1}{7} ln ∣ e^{x} - 6 ∣ + C

Solution steps

\int \frac{e ^{x}}{( e ^{x} - 6 ) ( e ^{x} + 1 )} d x

Expand

(e^{x} - 6) (e^{x} + 1) : e^{2 x} - 5 e^{x} - 6

= \int \frac{e ^{x}}{e ^{2 x} - 5 e ^{x} - 6} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 5 u - 6} d u

= \int \frac{1}{u ^{2} - 5 u - 6} d u

Take the partial fraction of

\frac{1}{u ^{2} - 5 u - 6} : - \frac{1}{7 ( u + 1 )} + \frac{1}{7 ( u - 6 )}

= \int - \frac{1}{7 ( u + 1 )} + \frac{1}{7 ( u - 6 )} 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}{7 ( u + 1 )} d u + \int \frac{1}{7 ( u - 6 )} d u

\int \frac{1}{7 ( u + 1 )} d u = \frac{1}{7} ln ∣ u + 1 ∣

\int \frac{1}{7 ( u - 6 )} d u = \frac{1}{7} ln ∣ u - 6 ∣

= - \frac{1}{7} ln ∣ u + 1 ∣ + \frac{1}{7} ln ∣ u - 6 ∣

Substitute back

u = e^{x}

= - \frac{1}{7} ln ∣ e^{x} + 1 ∣ + \frac{1}{7} ln ∣ e^{x} - 6 ∣

Add a constant to the solution

= - \frac{1}{7} ln ∣ e^{x} + 1 ∣ + \frac{1}{7} ln ∣ e^{x} - 6 ∣ + C

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What is the integral of (e^x)/((e^x-6)(e^x+1)) ?
The integral of (e^x)/((e^x-6)(e^x+1)) is -1/7 ln|e^x+1|+1/7 ln|e^x-6|+C