What is the integral of (e^x)/((e^{2x)+1)(e^x-3)} ?

The integral of (e^x)/((e^{2x)+1)(e^x-3)} is 1/10 (-1/2 ln|e^{2x}+1|-3arctan(e^x))+1/10 ln|e^x-3|+C

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integral of (e^x)/((e^{2x)+1)(e^x-3)}

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

\frac{1}{10} (- \frac{1}{2} ln e^{2 x} + 1 - 3 arctan (e^{x})) + \frac{1}{10} ln ∣ e^{x} - 3 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{( u ^{2} + 1 ) ( u - 3 )} : \frac{- u - 3}{10 ( u ^{2} + 1 )} + \frac{1}{10 ( u - 3 )}

= \int \frac{- u - 3}{10 ( u ^{2} + 1 )} + \frac{1}{10 ( u - 3 )} 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{- u - 3}{10 ( u ^{2} + 1 )} d u + \int \frac{1}{10 ( u - 3 )} d u

\int \frac{- u - 3}{10 ( u ^{2} + 1 )} d u = \frac{1}{10} (- \frac{1}{2} ln u^{2} + 1 - 3 arctan (u))

\int \frac{1}{10 ( u - 3 )} d u = \frac{1}{10} ln ∣ u - 3 ∣

= \frac{1}{10} (- \frac{1}{2} ln u^{2} + 1 - 3 arctan (u)) + \frac{1}{10} ln ∣ u - 3 ∣

Substitute back

u = e^{x}

= \frac{1}{10} (- \frac{1}{2} ln (e^{x})^{2} + 1 - 3 arctan (e^{x})) + \frac{1}{10} ln ∣ e^{x} - 3 ∣

Apply exponent rule:

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

= \frac{1}{10} (- \frac{1}{2} ln e^{2 x} + 1 - 3 arctan (e^{x})) + \frac{1}{10} ln ∣ e^{x} - 3 ∣

Add a constant to the solution

= \frac{1}{10} (- \frac{1}{2} ln e^{2 x} + 1 - 3 arctan (e^{x})) + \frac{1}{10} ln ∣ e^{x} - 3 ∣ + C

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What is the integral of (e^x)/((e^{2x)+1)(e^x-3)} ?
The integral of (e^x)/((e^{2x)+1)(e^x-3)} is 1/10 (-1/2 ln|e^{2x}+1|-3arctan(e^x))+1/10 ln|e^x-3|+C