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How you will evaluate ∫logxdx.
Asked by Rahul Srivastava Last Modified
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.
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Now take
as first function and 1 as second function.
then integrate by parts.
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Sreelakshmi S
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 less
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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