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(C) To evaluate the integral $I = \int (\log x)^2 dx$, we use the method of integration by parts: $\int u dv = uv - \int v du$. Let $u = (\log x)^2$ a

Vedclass · Accepted Answer

$x(\log x)^2 - 2x(\log x - 1) + c$

Vedclass · Answer

(C) To evaluate the integral $I = \int (\log x)^2 dx$, we use the method of integration by parts: $\int u dv = uv - \int v du$. Let $u = (\log x)^2$ and $dv = dx$. Then $du = 2 \log x \cdot \frac{1}{x} dx$ and $v = x$. Applying the formula: $I = x(\log x)^2 - \int x \cdot \frac{2 \log x}{x} dx + c$ $I = x(\log x)^2 - 2 \int \log x dx + c$ Now, integrate $\int \log x dx$ using integration by parts again (let $u = \log x, dv = dx$): $\int \log x dx = x \log x - \int x \cdot \frac{1}{x} dx = x \log x - x$. Substituting this back into the expression for $I$: $I = x(\log x)^2 - 2(x \log x - x) + c$ $I = x(\log x)^2 - 2x \log x + 2x + c$ $I = x(\log x)^2 - 2x(\log x - 1) + c$.

$\int (\log x)^2 dx =$

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