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Question

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

Vedclass · Accepted Answer

$\frac{x^2}{2} \log x - \frac{x^2}{4} + c$

Vedclass · Answer

(B) To evaluate the integral $\int x \log x \, dx$,we use the integration by parts formula: $\int u \, dv = uv - \int v \, du$.Let $u = \log x$ and $dv = x \, dx$.Then $du = \frac{1}{x} \, dx$ and $v = \int x \, dx = \frac{x^2}{2}$.Applying the formula:$\int x \log x \, dx = (\log x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx + c$$= \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx + c$$= \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} + c$$= \frac{x^2}{2} \log x - \frac{x^2}{4} + c$.

$\int x \log x \, dx = $

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