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The integral $∫logxdx$ is equal to : |
$xlog x + x + c$ $-xlog x - x + c$ $xlog x - x + c$ $xlog x + x^2 + c$ |
$xlog x - x + c$ |
The correct answer is Option (3) → $x\log x - x + c$ $I=\int\log xdx$ $=x\log x-\int\left(\frac{d}{dx}(\log x)\right)x\,dx$ $=x\log x-\int\frac{x}{x}dx$ $=x\log x-x+c$ |
