Why is the square of X equal to e ^ x * LNX

Why is the square of X equal to e ^ x * LNX

According to the properties of logarithm and exponent
X = e ^ (LNX) and ln (x ^ x) = x * LNX
So that's what you said
x^x=e^(lnx^x)=e^(x*lnx)

Finding indefinite integral LNX / 1 + x square DX numerator LNX denominator 1 plus x square
Finding the indefinite integral LNX / 1 + x square DX
Numerator LNX denominator 1 plus x squared

Flnx / (1 + x ^ 2) DX = fxlnx / (LNX + x ^ 2lnx) DX = 1 / 2fd (LNX + x ^ 2lnx) / (LNX + x ^ 2lnx) - 2F (1 / x + x) / (LNX + x ^ 2lnx) DX = 1 / 2ln | LNX + x ^ 2lnx | - 2fdx / xlnx = 1 / 2ln | LNX + x ^ 2lnx | - 2fdlnx / LNX = 1 / 2ln | LNX + x ^ 2lnx | - 2

Find the indefinite integral of (LNX) square divided by {x times [1 + (LNX) square]},

Add up the integral with
∫ (LNX) square divided by {x times [1 + (LNX) square]} DX
=∫(lnx)^2/[1+(lnx)^2]}dlnx
=∫{1-1/[1+(lnx)^2]}dlnx
=lnx-arctanlnx+C

What is the original function of xlnx
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