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HallsofIvy
#2
Feb28-12, 07:27 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,552
No, amazingly, you are wrong and your teacher is right!

Since that integrand does not exist at x= 0, you have to use the definition:
[tex]\int_{-1}^1 \frac{dx}{x}= \lim_{\alpha\to 0^-}\int_{-1}^\alpha \frac{dx}{x}+ \lim_{\beta\to 0^+}\int_\beta^1\frac{dx}{x}[/tex]
and those limits do not exist.
You cannot just evaluate the anti-derivative at 1, -1, and then subtract- that's ignoring the whole problem of what happens at 1.