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

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

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

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

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

Solution steps

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

Factor

e^{x} - 4 e^{- x} : e^{- x} (e^{x} + 2) (e^{x} - 2)

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Simplify

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

= \int \frac{e ^{x}}{( e ^{x} + 2 ) ( e ^{x} - 2 )} d x

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{( u + 2 ) ( u - 2 )} : - \frac{1}{4 ( u + 2 )} + \frac{1}{4 ( u - 2 )}

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

\int \frac{1}{4 ( u + 2 )} d u = \frac{1}{4} ln ∣ u + 2 ∣

\int \frac{1}{4 ( u - 2 )} d u = \frac{1}{4} ln ∣ u - 2 ∣

= - \frac{1}{4} ln ∣ u + 2 ∣ + \frac{1}{4} ln ∣ u - 2 ∣

Substitute back

u = e^{x}

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

Add a constant to the solution

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

y^{^{'}} = \frac{( y ^{4} - 1 )}{4 x y ^{3}}

\int \frac{8}{w ^{2} 4 - w ^{2}} d w

\frac{d}{d x} (x^{4} - 3)

\frac{d y}{d x} = e^{- a \cdot x} - b \cdot y, y (0) = 0

\int \frac{( 2 r + 1 )}{r ^{2} - 3 r + 2} d r

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