Question

\(\int \frac{8^{1+x} + 4^{1+x}}{2^{2x}} dx =\):

• A. \(\frac{2^{x}}{\log 2} + 4x + C\)
• B. \(8 \cdot \frac{2^{x}}{\log 2} - 4x + C\)
• C. \(8 \cdot \frac{2^{x}}{\log 2} + 4x + C\)
• D. \(\frac{2^{x}}{\log 2} - 4x + C\)

Hint:

For integrals involving exponential functions like \( a^{x} \), use the formula \(\int a^{x} dx = \frac{a^{x}}{\ln a} + C\), and simplify expressions by converting to a common base.

Answer 1

The Correct Option is C

Step 1: Simplify the integrand.
Rewrite the terms using base 2:
\[ 8^{1+x} = (2^3)^{1+x} = 2^{3(1+x)} = 2^{3 + 3x}, \quad 4^{1+x} = (2^2)^{1+x} = 2^{2(1+x)} = 2^{2 + 2x}, \quad 2^{2x} = (2^2)^x = 2^{2x} \] \[ \frac{8^{1+x} + 4^{1+x}}{2^{2x}} = \frac{2^{3 + 3x} + 2^{2 + 2x}}{2^{2x}} = 2^{3 + 3x - 2x} + 2^{2 + 2x - 2x} = 2^{3 + x} + 2^2 = 2^3 \cdot 2^x + 4 = 8 \cdot 2^x + 4 \] So the integral becomes: \[ \int (8 \cdot 2^x + 4) dx \]
Step 2: Integrate term by term.
- First term: \(\int 8 \cdot 2^x dx\):
\[ \int 8 \cdot 2^x dx = 8 \int 2^x dx = 8 \int e^{x \ln 2} dx = 8 \cdot \frac{e^{x \ln 2}}{\ln 2} = 8 \cdot \frac{2^x}{\ln 2} \] (Note: \(\ln 2 = \log_e 2\), and the options use \(\log 2\), which is interpreted as the natural logarithm in this context, consistent with calculus conventions.)
- Second term: \(\int 4 dx\):
\[ \int 4 dx = 4x + C_2 \] Combine: \[ \int (8 \cdot 2^x + 4) dx = 8 \cdot \frac{2^x}{\ln 2} + 4x + C \]
Step 3: Match with the options.
The result is: \[ 8 \cdot \frac{2^x}{\log 2} + 4x + C \] This matches option (3) exactly, confirming the solution is correct. Final Answer: \[ \boxed{3} \]