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$$
\int\left(\frac{8^{1+x}+4^{1+x}}{2^{2 x}}\right) d x=
$$
Options:
\int\left(\frac{8^{1+x}+4^{1+x}}{2^{2 x}}\right) d x=
$$
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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