$\int \frac{e^{5 x} + e^{3 x}}{e^{x} + e^{- x}} d x =$

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Solution

$I = \int \frac{e^{5 x} + e^{3 x}}{e^{x} + e^{- x}} d x = \int \frac{e^{5 x} + e^{3 x}}{e^{2 x} + 1} e^{x} d x$
Put $t = e^{x} ⟹ d t = e^{x} d x$
$I = \int \frac{t^{5} + t^{3}}{t^{2} + 1} d t = \int \frac{t^{3} (t^{2} + 1)}{(t^{2} + 1)} d t = \int t^{3} d t = \frac{t^{4}}{4} + c = \frac{e^{4 x}}{4} + C$

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