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How you will evaluate ∫logxdx.

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CBSE/Class 11

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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 more

∫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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1 Comments

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. read more

Now take

 as first function and 1 as second function.

then integrate by parts.

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1 Comments

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. read more

∫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
1 Comments

Build Yourself For The Future

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 &... read more

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

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1 Comments

IIM_home_Tuition Tutor Chennai mathematics 100%

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

Tutor with 2 years experience in teaching - Maths, English, Social, EVS, Science

The answer to this is 1/x.
Comments

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.
Comments

I will try to give my best but its totally depends on you thanks

I= x logx -x +C
Comments

Experienced Teacher, Tutor and Trainer with 6 years of experience in teaching

Take log x as log x. 1 and use integration by parts ie product rule you will get x log x - x + c.
Comments

Mathematics professional with 4year experience

Just apply ILATE formula and consider it as 1×logx.
Comments

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