What is the integral of sqrt(ln(x))

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SUMMARY

The integral of sqrt(ln(x)) can be approached using substitutions to simplify the expression. First, substitute y = ln(x), followed by y = z², leading to the integral ∫ z² e^(z²) dz. This integral can be evaluated using the error function, specifically the erfi function, which is necessary for functions that do not have elementary integrals. The final result is expressed as ½(√π*i* erf(i√(ln(x))) + 2x√(ln(x))) + C.

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what is the integral of sqrt(ln(x))
NOTE:(I do not need the final answer,I know it. I want to Know how to solve it.)
 
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We usually discourage such terse questions... especially without the poster any showing to attempts at the solution (as you've done in several recent posts).
Can you tell us your answer and how are approaching arriving at it?
 
Do the substitution y=lnx, and then the substitution y=z^2. You ll end up with an integral of the form \int z^2e^{z^2}dz which if you look up at wolfram i believe you ll understand how it is calculated (at first glance the integral would be equal to ze^{z^2}/2-f(z) where f(z) should be such that f'(z)=e^{z^2}/2). Such a function f(z) is the erfi function.
 
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Look i substituted (x=e^(-t^2))
dx=-2t*e^(-t^2)dt

∫√(ln(e^(-t^2))*-2t*e^(-t^2)dt
-2∫√(-t^2)*t*e^(-t^2)dt

-2∫it^2*e^(-t^2)dt
When i arrived here i stopped.
but i used integral calculator to find the final answer.

this is the final answer:

½(√π*i* erf(i√(ln(x)))+2x√(ln(x)))+C
 
Do you understand what that means? Do you understand that almost all such functions have NO integral in terms of elementary functions?
 
HallsofIvy said:
Do you understand what that means? Do you understand that almost all such functions have NO integral in terms of elementary functions?
yes i understand. thank you.
 

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