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How do you evaluate the integral #int lnx/xdx# from 0 to 1 if it converges?

Answer 1

Aurora Alvarado

Does not converge

#int_0^1 lnx/xdx#

#=int_0^1 d/dx (1/2 ln^2x)dx#

#= 1/2 lim_(t to 0) [ ln^2x ]_t^1#

#= - 1/2 lim_(t to 0) ln^2t#

and limit of product is product of limits

#= - 1/2 lim_(t to 0) lnt * lim_(t to 0) lnt#

Given #lim_(t to 0) lnt = -oo#, does not converge

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Answer 2

Luke Alvarez

To evaluate the integral ( \int_0^1 \frac{\ln(x)}{x} , dx ), you can use integration by parts.

Let ( u = \ln(x) ) and ( dv = \frac{1}{x} , dx ).
Calculate the differentials ( du ) and ( v ).
Apply the integration by parts formula: [ \int u , dv = uv - \int v , du ]
Substitute the values of ( u ), ( v ), ( du ), and ( dv ) into the formula.
Evaluate the integral from the resulting equation.

After following these steps, you should obtain the value of the integral from 0 to 1 if it converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.