int log_{10} x , dx = | vedclass

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(D) We know that $\log_{10} x = \frac{\log_e x}{\log_e 10}$.Thus,$\int \log_{10} x \, dx = \int \frac{\log_e x}{\log_e 10} \, dx = \frac{1}{\log_e 10}

Vedclass · Accepted Answer

$x(\log_{10} x - \log_{10} e) + c$

Vedclass · Answer

(D) We know that $\log_{10} x = \frac{\log_e x}{\log_e 10}$.Thus,$\int \log_{10} x \, dx = \int \frac{\log_e x}{\log_e 10} \, dx = \frac{1}{\log_e 10} \int \log_e x \, dx$.Using integration by parts,$\int \log_e x \, dx = x \log_e x - x + c$.Therefore,$\int \log_{10} x \, dx = \frac{1}{\log_e 10} (x \log_e x - x) + c$.$= \frac{x \log_e x}{\log_e 10} - \frac{x}{\log_e 10} + c$.Since $\frac{\log_e x}{\log_e 10} = \log_{10} x$ and $\frac{1}{\log_e 10} = \log_{10} e$,we get:$= x \log_{10} x - x \log_{10} e + c = x(\log_{10} x - \log_{10} e) + c$.

$\int \log_{10} x \, dx = $

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