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

The integral of (e^{-x})/(e^x-5+4e^{-x)} is 1/4 x-1/3 ln|e^x-1|+1/12 ln|e^x-4|+C

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

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

\frac{1}{4} x - \frac{1}{3} ln ∣ e^{x} - 1 ∣ + \frac{1}{12} ln ∣ e^{x} - 4 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u ^{2} - 5 u + 4 )} d u

= \int \frac{1}{u ( u ^{2} - 5 u + 4 )} d u

Take the partial fraction of

\frac{1}{u ( u ^{2} - 5 u + 4 )} : \frac{1}{4 u} - \frac{1}{3 ( u - 1 )} + \frac{1}{12 ( u - 4 )}

= \int \frac{1}{4 u} - \frac{1}{3 ( u - 1 )} + \frac{1}{12 ( 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}{4 u} d u - \int \frac{1}{3 ( u - 1 )} d u + \int \frac{1}{12 ( u - 4 )} d u

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{3 ( u - 1 )} d u = \frac{1}{3} ln ∣ u - 1 ∣

\int \frac{1}{12 ( u - 4 )} d u = \frac{1}{12} ln ∣ u - 4 ∣

= \frac{1}{4} ln ∣ u ∣ - \frac{1}{3} ln ∣ u - 1 ∣ + \frac{1}{12} ln ∣ u - 4 ∣

Substitute back

u = e^{x}

= \frac{1}{4} ln ∣ e^{x} ∣ - \frac{1}{3} ln ∣ e^{x} - 1 ∣ + \frac{1}{12} ln ∣ e^{x} - 4 ∣

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

= \frac{1}{4} x - \frac{1}{3} ln ∣ e^{x} - 1 ∣ + \frac{1}{12} ln ∣ e^{x} - 4 ∣

Add a constant to the solution

= \frac{1}{4} x - \frac{1}{3} ln ∣ e^{x} - 1 ∣ + \frac{1}{12} ln ∣ e^{x} - 4 ∣ + C

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What is the integral of (e^{-x})/(e^x-5+4e^{-x)} ?
The integral of (e^{-x})/(e^x-5+4e^{-x)} is 1/4 x-1/3 ln|e^x-1|+1/12 ln|e^x-4|+C