Question

Solve:
$\int \frac{8^{1+x}+4^{1-x}}{2^x}$

Answer 1

$\int \frac{8^{1+x}+4^{1-x}}{2^x}dx$

$=\int \frac{8{\left(2^x\right)}^3+4\cdot 2^{-2x}}{2^x}$

$=\int \frac{2^x(8{\left(2^x\right)}^2+4{\left(2^x\right)}^{-3})}{2^x}$

$=8\int 4^{x}dx+4\int 8^{-x}dx$

$=8\left(\frac{4^x}{\log 4}\right)+4\left(\frac{-8^{-x}}{\log 8}\right)+c$

$=\frac{2^3\cdot 2^{2x}}{2\log 2}-\frac{2^2\cdot 2^{-3x}}{3\log 2}+c$

$=\frac{3\cdot 2^{3+2x}-2^{3(1-x)}}{6\log 2}+c$