If ∫log10x dx=K.xlogf(x)+c, then K,f(x)=

I=∫log10x dx=log10e∫logex dx=log10e[xlogx−∫x.1xdx]+c=log10e[xlogx−x]+c=log10e.xlog(xe)+c∴K=log10e,f(x)=xe

If ∫log10x dx=K.xlogf(x)+c, then K,f(x)=

I=∫log10x dx=log10e∫logex dx=log10e[xlogx−∫x.1xdx]+c=log10e[xlogx−x]+c=log10e.xlog(xe)+c∴K=log10e,f(x)=xe

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If $\int \log_{10} x d x = K . x \log f (x) + c,$ then $K, f (x) =$

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$\log_{10} e, \frac{x}{e}$

$\log_{10} e, \frac{e}{x}$

$\log_{2} 10, \frac{x}{e}$

$\log_{e} 10, \frac{x}{e}$

answer is D.

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Detailed Solution

$I = \int \log_{10} x d x$

$= \log_{10} e \int \log_{e} x d x$

$= \log_{10} e [x \log x - \int x . \frac{1}{x} d x] + c$

$= \log_{10} e [x \log x - x] + c$

$= \log_{10} e . x \log (\frac{x}{e}) + c$

$∴ K = \log_{10} e, f (x) = \frac{x}{e}$

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If $\int \log_{10} x d x = K . x \log f (x) + c,$ then $K, f (x) =$