How you will evaluate∫logxdx.

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∫log x dx = x log x -x Proof : Using integration by parts, ∫udv = uv - ∫vdu In ∫ log x dx, take, u=logx => du= (1/x) . dx ∫dv=∫dx => v=x Now substituting, ∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C where C is constant.
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∫log x dx = x log x -x Proof : Using integration by parts, ∫udv = uv - ∫vdu In ∫ log x dx, take, u=logx => du= (1/x) . dx ∫dv=∫dx => v=x Now substituting, ∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C where C is constant. read less
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1st Class in M.Com(Finance) from CU. Trying to teach young minds for better future.

Now take as first function and 1 as second function. then integrate by parts.
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Now take as first function and 1 assecond function. then integrate by parts. read less
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Biotechnology graduate and Medical Coder, experienced tutor for Science subjects and Mathematics.

∫log x dx = x log x -x Proof : Using integration by parts, ∫udv = uv - ∫vdu In ∫ log x dx, take, u=logx => du= (1/x) . dx ∫dv=∫dx => v=x Now substituting, ∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C where C is constant.
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∫log x dx = x log x -x Proof : Using integration by parts, ∫udv = uv - ∫vdu In ∫ log x dx, take, u=logx => du= (1/x) . dx ∫dv=∫dx => v=x Now substituting, ∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C where C is constant. read less
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Using ILATE RULE where I stands for inverse trigonometric functions L stands for logarithmic functions A stands for algebraic functions T stands for trigonometric functions E stands for exponential functions. Solve this integration using integration by part where consider logx as first function &...
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Using ILATE RULE where I stands for inverse trigonometric functions L stands for logarithmic functions A stands for algebraic functions T stands for trigonometric functions E stands for exponential functions. Solve this integration using integration by part where consider logx as first function & 1as second function. Ans- xlogx-x read less
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I will try to give my best but its totally depends on you thanks

I= x logx -x +C
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An internationally certified soft skills trainer

Take log x as log x. 1 and use integration by parts ie product rule you will get x log x - x + c.
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Mathematics professional with 4year experience

Just apply ILATE formula and consider it as 1×logx.
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Tutor with 2 years experience in teaching - Maths, English, Social, EVS, Science

The answer to this is 1/x.
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NPTEL Gold Medalist Mechanical Engineering Graduate

Apply the formula of integration by parts .Assume log x as first function and 1 as second function. Write 1 as x raised to the power zero.
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IIM_home_Tuition Tutor Chennai mathematics 100%

It's come from basic formula 1/x+c c is constant.
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