Q.$\dfrac{8^{1+x}+4^{1-x}}{2^{x}}$

Step 1. Rewrite $8^{1+x}$. $8^{1+x}=8\cdot8^x=8\cdot2^{3x}$; dividing by $2^x$ gives $8\cdot2^{2x}=8\cdot4^x$.

Step 2. Rewrite $4^{1-x}$. $4^{1-x}=4\cdot4^{-x}=4\cdot2^{-2x}$; dividing by $2^x$ gives $4\cdot2^{-3x}=4\cdot8^{-x}$.

Step 3. So the integrand is $8\cdot4^x+4\cdot8^{-x}$.

Step 4. Integrate each exponential. $\displaystyle\int8\cdot4^xdx=\dfrac{8\cdot4^x}{\log4}$, $\displaystyle\int4\cdot8^{-x}dx=\dfrac{4\cdot8^{-x}}{-\log8}=-\dfrac{4\cdot8^{-x}}{\log8}$. …

• Sign error on the derivative of $8^{-x}$ (needs an extra $-\log8$) …