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Question

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

Vedclass · Accepted Answer

$\frac{x^{2}}{4}(2 \log x - 1) + c$

Vedclass · Answer

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

$\int x \log x \, dx$ is equal to

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