Evaluate the integral: int left(frac{8^{1+x}+ | vedclass

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(C) Given the integral: $I = \int \left(\frac{8^{1+x}+4^{1+x}}{2^{2x}}\right) dx$ We can rewrite the terms as: $8^{1+x} = 8 \cdot 8^x = 8 \cdot (2^3)^

Vedclass · Accepted Answer

$8 \cdot \frac{2^x}{\log 2}+4x+C$

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(C) Given the integral: $I = \int \left(\frac{8^{1+x}+4^{1+x}}{2^{2x}}\right) dx$ We can rewrite the terms as: $8^{1+x} = 8 \cdot 8^x = 8 \cdot (2^3)^x = 8 \cdot 2^{3x}$ and $4^{1+x} = 4 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x}$ Substituting these into the integral: $I = \int \frac{8 \cdot 2^{3x} + 4 \cdot 2^{2x}}{2^{2x}} dx$ Dividing each term by $2^{2x}$: $I = \int (8 \cdot 2^{3x-2x} + 4) dx = \int (8 \cdot 2^x + 4) dx$ Integrating with respect to $x$: $I = 8 \int 2^x dx + \int 4 dx$ Using the formula $\int a^x dx = \frac{a^x}{\log_e a} + C$: $I = 8 \cdot \frac{2^x}{\log_e 2} + 4x + C$

Evaluate the integral: $\int \left(\frac{8^{1+x}+4^{1+x}}{2^{2x}}\right) dx$

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