Question
$\int \frac{ e ^{2 x}+ e ^{-2 x}}{ e ^x} d x$

Answer

$\int \frac{ e ^{2 x}+ e ^{-2 x}}{ e ^x} d x=\int\left( e ^x+ e ^{-3 x}\right) d x$
$= e ^x-\frac{ e ^{-3 x}}{3}+ c$
$= e ^x-\frac{1}{3 e ^{3 x}}+ c $

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