If ∫log⁡(log$$⁡$$x)+1(log$$⁡$$x)2dx=x$$[$$f(x)−$$g(x)$$]$$+C$$, then

Correct option is 1f(x)=log⁡(log⁡x);g(x)=1log⁡x

Questions

If $\int [\log (\log x) + \frac{1}{(\log x)^{2}}] d x$
$= x [f (x) - g (x)] + C$ , then

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f(x)=log⁡(log⁡x);g(x)=1log⁡x

f(x)=log⁡x;g(x)=1log⁡x

f(x)=1log⁡x;g(x)=log⁡(log⁡x)

f(x)=1xlog⁡x;g(x)=1log⁡x

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detailed solution

Correct option is A

∫log⁡(log⁡x)dx=xlog⁡(log⁡x)−∫x1log⁡x⋅1xdx=xlog⁡(log⁡x)−∫dxlog⁡x=xlog⁡(log⁡x)−x1log⁡x+∫x(−1)(log⁡x)21xdx⇒∫log⁡(log⁡x)+1(log⁡x)2dx=x[f(x)−g(x)]+Cwhere f(x)=log⁡(log⁡x)and g(x)=1log⁡x

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If ∫log⁡(log⁡x)+1(log⁡x)2dx=x[f(x)−g(x)]+C, then

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

If $\int [\log (\log x) + \frac{1}{(\log x)^{2}}] d x$
$= x [f (x) - g (x)] + C$ , then