What is the integral of (e^x)/((e^x-5)(e^x-11)) ?

The integral of (e^x)/((e^x-5)(e^x-11)) is -1/6 ln|e^x-5|+1/6 ln|e^x-11|+C

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integral of (e^x)/((e^x-5)(e^x-11))

\int \frac{e ^{x}}{( e ^{x} - 5 ) ( e ^{x} - 11 )} d x

- \frac{1}{6} ln ∣ e^{x} - 5 ∣ + \frac{1}{6} ln ∣ e^{x} - 11 ∣ + C

Solution steps

\int \frac{e ^{x}}{( e ^{x} - 5 ) ( e ^{x} - 11 )} d x

Apply u-substitution

: \int \frac{1}{( u - 5 ) ( u - 11 )} d u

= \int \frac{1}{( u - 5 ) ( u - 11 )} d u

Take the partial fraction of

\frac{1}{( u - 5 ) ( u - 11 )} : - \frac{1}{6 ( u - 5 )} + \frac{1}{6 ( u - 11 )}

= \int - \frac{1}{6 ( u - 5 )} + \frac{1}{6 ( u - 11 )} 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}{6 ( u - 5 )} d u + \int \frac{1}{6 ( u - 11 )} d u

\int \frac{1}{6 ( u - 5 )} d u = \frac{1}{6} ln ∣ u - 5 ∣

\int \frac{1}{6 ( u - 11 )} d u = \frac{1}{6} ln ∣ u - 11 ∣

= - \frac{1}{6} ln ∣ u - 5 ∣ + \frac{1}{6} ln ∣ u - 11 ∣

Substitute back

u = e^{x}

= - \frac{1}{6} ln ∣ e^{x} - 5 ∣ + \frac{1}{6} ln ∣ e^{x} - 11 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ e^{x} - 5 ∣ + \frac{1}{6} ln ∣ e^{x} - 11 ∣ + C

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What is the integral of (e^x)/((e^x-5)(e^x-11)) ?
The integral of (e^x)/((e^x-5)(e^x-11)) is -1/6 ln|e^x-5|+1/6 ln|e^x-11|+C