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Q.The solution of dydx+yxlog⁡x=1x\dfrac{dy}{dx} + \dfrac{y}{x \log x} = \dfrac{1}{x} is

a
ylog⁡x=log⁡x+Cy \log x = \log x + C
b
ylog⁡(log⁡x)=log⁡x+Cy \log(\log x) = \log x + C
c
y=log⁡(log⁡x)+Cy = \log(\log x) + C
d
y=log⁡x+Clog⁡(log⁡x)y = \log x + C \log(\log x)

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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