Answer
$$ \displaystyle\int \dfrac{{8}^{1+x}+{4}^{1-x}}{{2}^{x}}\, \mathrm d x = {{4^{-{{\ln 2\,x}\over{\ln 4}}-x+1}\,\left(\ln 4\,e^{2\,\ln 2\, x}\,4^{{{\ln 2\,x}\over{\ln 4}}+x}+\ln 2\,e^{2\,\ln 2\,x}\,4^{{{ \ln 2\,x}\over{\ln 4}}+x}-\ln 2\right)}\over{\ln 2\,\left(\ln 4 +\ln 2\right)}} $$
Explanation
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