Answer
$$\frac{1}{3}{e^{3x}} - {e^x} - {e^{ - x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{4x}} - {e^{2x}} + 1}}{{{e^x}}}} dx \cr
& {\text{Distributive property}} \cr
& = \int {\left( {\frac{{{e^{4x}}}}{{{e^x}}} - \frac{{{e^{2x}}}}{{{e^x}}} + \frac{1}{{{e^x}}}} \right)} dx \cr
& {\text{Use the law }}\frac{{{e^a}}}{{{e^b}}} = {e^{a - b}} \cr
& = \int {\left( {{e^{4x - x}} - {e^{2x - x}} + {e^{ - x}}} \right)} dx \cr
& = \int {\left( {{e^{3x}} - {e^x} + {e^{ - x}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{3}{e^{3x}} - {e^x} - {e^{ - x}} + C \cr} $$
