∫2log4xdx=∫2log2x1dx=∫212log2xdx ∵logaxxm=mnlogax=∫2log2xdx=∫x dx (∵alogax=x)=2x3/23+c

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

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$\frac{1}{2 \sqrt{x}} + c$

$\sqrt{x} + c$

$\frac{2}{3} x^{3 / 2} + c$

$- \frac{1}{2 \sqrt{x}} + c$

answer is B.

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Detailed Solution

$\int 2^{\log_{4} x} d x = \int 2^{\log_{2}^{x^{1}}} d x$

$= \int 2^{\frac{1}{2} \log_{2}^{x}} d x (∵ \log_{a^{x}}^{x^{m}} = \frac{m}{n} \log_{a}^{x})$

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

$= \int \sqrt{x} d x (∵ a^{\log_{a}^{x}} = x)$

$= \frac{2 x^{3 / 2}}{3} + c$

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