What is the integral of (e^x)/((e^x-1)(e^{2x)+1)} ?

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

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integral of (e^x)/((e^x-1)(e^{2x)+1)}

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

\frac{1}{2} ln ∣ e^{x} - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln e^{2 x} + 1 - arctan (e^{x})) + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

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

= \frac{1}{2} ln ∣ u - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 - arctan (u))

Substitute back

u = e^{x}

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

Add a constant to the solution

= \frac{1}{2} ln ∣ e^{x} - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln e^{2 x} + 1 - arctan (e^{x})) + C

\frac{x}{x ^{2} + 64}

x \to 3 lim (\frac{( x ^{2} - 6 )}{( x - 9 )})

\frac{d}{d x} (lo g_{6} (3 x^{3} + 4 x^{2} + 4 x))

\int 64 - 5 x^{2} d x

\int \frac{( x ^{2} )}{20 - 4 x + x ^{2}} d x

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