question_answer

Find \[\int{\frac{{{8}^{1\,+\,x}}+{{4}^{1\,-\,x}}}{{{2}^{x}}}\,}dx.\]

Answer:

Let \[l=\int{\frac{{{8}^{1\,+\,x}}+{{4}^{1\,-\,x}}}{{{2}^{x}}}\,}dx\] \[\Rightarrow \] \[l=\int{\frac{{{2}^{3x+3}}+{{2}^{2-2x}}}{{{2}^{x}}}\,}dx=\int{\left( \frac{{{2}^{3x+3}}}{{{2}^{x}}}+\frac{{{2}^{2x-2}}}{{{2}^{x}}} \right)}\,dx\] \[\Rightarrow \] \[=\int{({{2}^{2x+3}}+{{2}^{2-3x}})}\,dx=\frac{{{2}^{2x+3}}}{2\,\log 2}+\frac{{{2}^{2-3x}}}{(-\,3)log2}+C\]             \[=\frac{1}{\log }\left[ {{2}^{2x\,+\,2}}-\frac{1}{3}\cdot {{2}^{2\,-\,3x}} \right]+C\]