What are you trying to do? The syntax you use depend on what you want to achieve.
Indefinite integral : Integral( ln(x),x) gives the correct answer xln(x)-x
Definite integral : Integral(ln(x),x,1,2) gives the integral of ln(x) between x=1 and x=2
Definite integral between 1 and y : Integral (ln(x), x,1,y) gives the correct expression yln(y)-y +1
By inserting the =0 you are in fact assigning the result of the integration to value 0, and the calculator calculates the integral and assigns 0 to it. This is what my TIVoyage 200 PLT gives me:
{Integral((lnx),x,1) = 0.} returns {x.ln(x)-x+1=0}
which is an equation.
Thanks
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