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Solution to a trigonometric problem requested by one student

B.sudhakar

15/01/2017 4 0

∫ 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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