show whether the integral converges or diverges is the integral finite? \[\int_1^\infty \frac{1}{2x - 1} \mathrm{d}x\]
diverges for sure
in order for it to converge, the degree of the denominator must be larger than the degree of the numerator by more than one
if you integrate that thing, you will get the log of something when you take the limit as \(x\to \infty\) of the log of whatever, it will go to infinity
\[log_{something} ( \inf ) = \inf\]
for example \[\int_1^{\infty}\frac{x+1}{x^2-2}dx\] diverges, whereas \[\int_1^{\infty}\frac{x+1}{x^3-2}dx\] converges
I see
\[anybase^{\inf} = \inf\]is what the logarithm x->inf is about?
^ "when you take the limit as x→∞ of the log of whatever, it will go to infinity"
\[\lim_{x \rightarrow \inf} log_{whatever} ( x) = \inf\]is the same statement as \[whatever^{\inf} = x,x\rightarrow \inf\]?
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