int [(log_{2} x)^2 + 2 log_{2} x] dx = | vedclass

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(C) Let $I = \int [(\log_{2} x)^2 + 2 \log_{2} x] dx$. We know that $\frac{d}{dx} [x(\log_{2} x)^2] = 1 \cdot (\log_{2} x)^2 + x \cdot 2(\log_{2} x) \

Vedclass · Accepted Answer

$x(\log_{2} x)^2 + c$

Vedclass · Answer

(C) Let $I = \int [(\log_{2} x)^2 + 2 \log_{2} x] dx$. We know that $\frac{d}{dx} [x(\log_{2} x)^2] = 1 \cdot (\log_{2} x)^2 + x \cdot 2(\log_{2} x) \cdot \frac{d}{dx}(\log_{2} x)$. Since $\frac{d}{dx}(\log_{2} x) = \frac{1}{x \ln 2}$,this does not simplify directly to the integrand. Let's use integration by parts on the first term: $\int (\log_{2} x)^2 dx$. Let $u = (\log_{2} x)^2$ and $dv = dx$. Then $du = 2(\log_{2} x) \cdot \frac{1}{x \ln 2} dx$ and $v = x$. $\int (\log_{2} x)^2 dx = x(\log_{2} x)^2 - \int x \cdot \frac{2 \log_{2} x}{x \ln 2} dx = x(\log_{2} x)^2 - \frac{2}{\ln 2} \int \log_{2} x dx$. This suggests the original problem likely assumes $\log x$ as the natural logarithm $\ln x$. Assuming $\log x = \ln x$: $I = \int [(\ln x)^2 + 2 \ln x] dx$. Using the derivative rule: $\frac{d}{dx} [x(\ln x)^2] = (\ln x)^2 + x \cdot 2 \ln x \cdot \frac{1}{x} = (\ln x)^2 + 2 \ln x$. Therefore,$\int [(\ln x)^2 + 2 \ln x] dx = x(\ln x)^2 + c$.

$\int [(\log_{2} x)^2 + 2 \log_{2} x] dx = $

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