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Question: Answered & Verified by Expert
$$
\int\left(\frac{8^{1+x}+4^{1+x}}{2^{2 x}}\right) d x=
$$
MathematicsDefinite IntegrationAP EAMCETAP EAMCET 2023 (18 May Shift 2)
Options:
  • A $\frac{2^x}{\log 2}+4 x+C$
  • B $8 \cdot \frac{2^x}{\log 2}-4 x+C$
  • C $8 \cdot \frac{2^{\mathrm{x}}}{\log 2}+4 \mathrm{x}+\mathrm{C}$
  • D $\frac{2^{\mathrm{x}}}{\log 2}-4 \mathrm{x}+\mathrm{C}$
Solution:
1258 Upvotes Verified Answer
The correct answer is: $8 \cdot \frac{2^{\mathrm{x}}}{\log 2}+4 \mathrm{x}+\mathrm{C}$
$\begin{aligned} & \text {} \int \frac{8^{1+\mathrm{x}}+4^{1+\mathrm{x}}}{2^{2 \mathrm{x}}}=\int \frac{8.8^{\mathrm{x}}+4 \cdot 4^{\mathrm{x}}}{2^{2 \mathrm{x}}} \mathrm{dx}+\mathrm{c} \\ & =\int \frac{8.2^{3 \mathrm{x}}+4.2^{2 \mathrm{x}}}{2^{2 \mathrm{x}}} \mathrm{dx}+\mathrm{c} \\ & =\int\left(8.2^{\mathrm{x}}+4\right) \mathrm{dx}+\mathrm{c}=8 \cdot \frac{2^{\mathrm{x}}}{\log _e 2}+4 \mathrm{x}+\mathrm{c}\end{aligned}$

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