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How you will evaluate ∫logxdx.

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Answer

∫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

1 Comments

∫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

1 Comments

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

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

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Comments

I= x logx -x +C

1

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