Cho \(F\left( x \right) = \left( {x - 1} \right){{...

C

- Hướng dẫn giải

Ta có \(\int {f\left( x \right).{{\rm{e}}^{2x}}{\rm{d}}x}  = \left( {x - 1} \right){{\rm{e}}^x} + C \Rightarrow f\left( x \right).{{\rm{e}}^{2x}} = {{\rm{e}}^x} + \left( {x - 1} \right).{{\rm{e}}^x} = x.{{\rm{e}}^x}\)

\( \Rightarrow f\left( x \right) = x.{{\rm{e}}^{ - x}} \Rightarrow f'\left( x \right) = \left( {1 - x} \right).{{\rm{e}}^{ - x}}\)

Suy ra \(\int {f'\left( x \right){{\rm{e}}^{2x}}} {\rm{d}}x = \int {\left( {1 - x} \right){{\rm{e}}^x}} {\rm{d}}x = \int {\left( {1 - x} \right){\rm{d}}\left( {{{\rm{e}}^x}} \right)}  = {{\rm{e}}^x}\left( {1 - x} \right) + \int {{{\rm{e}}^x}{\rm{d}}x}  = {{\rm{e}}^x}\left( {2 - x} \right) + C\).