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Question

(B) To evaluate the integral $\int \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = \log x$ and $d

Vedclass · Accepted Answer

$x \log x - x + c$

Vedclass · Answer

(B) To evaluate the integral $\int \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = \log x$ and $dv = dx$.Then,$du = \frac{1}{x} \, dx$ and $v = x$.Applying the formula:$\int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx + c$$= x \log x - \int 1 \, dx + c$$= x \log x - x + c$.Thus,the correct option is $B$.

$\int \log x \, dx = $

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