int frac{1}{log_x e} , dx = | vedclass

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(C) We know that $\log_x e = \frac{1}{\log_e x}$.Therefore,$\frac{1}{\log_x e} = \log_e x$.The integral becomes $\int \log_e x \, dx$.Using integratio

Vedclass · Accepted Answer

$x \log \left( \frac{x}{e} \right) + c$

Vedclass · Answer

(C) We know that $\log_x e = \frac{1}{\log_e x}$.Therefore,$\frac{1}{\log_x e} = \log_e x$.The integral becomes $\int \log_e x \, dx$.Using integration by parts,$\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$,where $u = \log_e x$ and $v = 1$.$\int \log_e x \cdot 1 \, dx = (\log_e x) \cdot x - \int \left( \frac{1}{x} \cdot x \right) \, dx$.$= x \log_e x - \int 1 \, dx = x \log_e x - x + c$.$= x(\log_e x - \log_e e) + c = x \log_e \left( \frac{x}{e} \right) + c$.

$\int \frac{1}{\log_x e} \, dx = $

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