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∫ log x dx use integration by parts. Let u = log x. ∴ du = 1/x dx. and. dv = dx ∴ v= x
By substituting u and dv in the formula ∫ u dv = u•v - ∫ v du
∫ log x dx = log x x - ∫ x 1/x dx ( x cancels out )
= x log x - ∫ dx
= x log x - x + C
∴ ∫ log x dx = x [ log x - 1 ] + C
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