∫2log42⁡2dx=∫2log22dx ∴logbnmm=mnlogba⁡alogax=x=∫212log22dx=∫2log2⁡x212dx=∫2log2xdx=∫xdx=x22+c

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$\int 2^{{\log_{4}^{x}}^{2}} d x =$

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$3 x^{2} + c$

$\frac{x^{2}}{2} + c$

$2 x^{2} + c$

$\frac{x^{2}}{4} + c$

answer is C.

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Detailed Solution

$\begin{array}{l} \int 2^{\log_{4}^{2} 2} d x \\ = \int 2^{\log_{2}^{2}} d x (∴ \log_{b^{n}}^{m^{m}} = \frac{m}{n} \log_{b}^{a} a^{\log_{a}^{x}} = x \\ = \int 2^{\frac{1}{2} \log_{2}^{2}} d x \\ = \int 2^{\log_{2} {(x^{2})}^{\frac{1}{2}}} d x = \int 2^{\log_{2}^{x}} d x \\ = \int x d x \\ = \frac{x^{2}}{2} + c \end{array}$