int log x^2 , dx = . . . . . . + C. | vedclass

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Question

(D) We need to evaluate the integral $I = \int \log x^2 \, dx$.Using the property of logarithms,$\log x^2 = 2 \log x$.So,$I = \int 2 \log x \, dx = 2

Vedclass · Accepted Answer

$2 x \log \left(\frac{x}{e}\right)$

Vedclass · Answer

(D) We need to evaluate the integral $I = \int \log x^2 \, dx$.Using the property of logarithms,$\log x^2 = 2 \log x$.So,$I = \int 2 \log x \, dx = 2 \int \log x \, dx$.Using integration by parts,$\int \log x \, dx = x \log x - x + C_1$.Therefore,$I = 2(x \log x - x) + C = 2x \log x - 2x + C$.We can rewrite this as $2x(\log x - 1) = 2x(\log x - \log e) = 2x \log \left(\frac{x}{e}\right) + C$.Thus,the correct option is $D$.

$\int \log x^2 \, dx =$ . . . . . . $+ C$.

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