$\int \frac{2 e^{5 x} + e^{4 x} - 4 e^{3 x} + 4 e^{2 x} + 2 e^{x}}{(e^{2 x} + 4) {(e^{2 x} - 1)}^{2}} d x =$

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$\tan^{- 1} \frac{e^{x}}{2} - \frac{1}{e^{2 x} - 1} + C$

$\tan^{- 1} e^{x} - \frac{1}{2 (e^{2 x} - 1)} + C$

$\tan^{- 1} \frac{e^{x}}{2} - \frac{1}{2 (e^{2 x} - 1)} + C$

$1 - \tan^{- 1} (\frac{e^{x}}{2}) + \frac{1}{2 (e^{2 x} - 1)} + C$

answer is C.

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Detailed Solution

$I = \int \frac{2 e^{5 x} + e^{4 x} - 4 e^{3 x} + 4 e^{2 x} + 2 e^{x}}{(e^{2 x} + 4) {(e^{2 x} - 1)}^{2}} dx$

$Put e^{x} = t$

$\begin{array}{l} ∴ I = \int \frac{(2 t^{4} - 4 t^{2} + 2)}{(t^{2} + 4) {(t^{2} - 1)}^{2}} dt + \int \frac{(t^{3} + 4 t) dt}{(t^{2} + 4) {(t^{2} - 1)}^{2}} \\ = 2 \int \frac{dt}{t^{2} + 4} + \frac{1}{2} \int \frac{2 tdt}{{(t^{2} - 1)}^{2}} \\ = \tan^{- 1} \frac{t}{2} - \frac{1}{2 (t^{2} - 1)} + C \\ = \tan^{- 1} \frac{e^{x}}{2} - \frac{1}{2 (e^{2 x} - 1)} + C \end{array}$