Tìm nguyên hàm F(x) của \[f\left( x \right) = \frac{{{2^x} - 1}}{{{e^x}}}.\] biết F(0)=1.

\[F\left( x \right) = \smallint \frac{{{2^x} - 1}}{{{e^x}}}dx = \smallint \left( {{2^x} - 1} \right){e^{ - x}}dx = \smallint {2^x}{e^{ - x}}dx - \smallint {e^{ - x}}dx\]

\[ = \smallint {2^x}{e^{ - x}}dx + {e^{ - x}} + {C_1} = I + {e^{ - x}} + {C_1}.\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = {2^x}}\\{dv = {e^{ - x}}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = {2^x}ln2dx}\\{v = - {e^{ - x}}}\end{array}} \right.\)

\[\begin{array}{l} \Rightarrow I = - {2^x}{e^{ - x}} + ln2\smallint {2^x}{e^{ - x}}dx + {C_2} = - {2^x}{e^{ - x}} + ln2.I + {C_2}\\ \Leftrightarrow (ln2 - 1)I + {C_2} = {2^x}{e^{ - x}} \Rightarrow I = \frac{{{2^x}{e^{ - x}}}} + {C_2}.\end{array}\]

\[ \Rightarrow F(x) = \frac{{{2^x}{e^{ - x}}}} + {e^{ - x}} + C = \frac{{{2^x}}}{{(ln2 - 1){e^x}}} + \frac{1}{{{e^x}}} + C\]

\[ \Rightarrow F(0) = \frac{1} + 1 + C = 1 \Rightarrow C = - \frac{1}\]

\[ \Rightarrow F(x) = \frac{{{2^x}}}{{(ln2 - 1){e^x}}} + \frac{1}{{{e^x}}} - \frac{1}\]

\[ = \frac{1}{\left( {\frac{2}{e}} \right)^x} + {\left( {\frac{1}{e}} \right)^x} - \frac{1}\]

Đáp án cần chọn là: B