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converge or diverge: from 1 to infinty: 1/(3x+1)^2dx
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@eliassaab medal? lol
convergent
please show how to do it
this one is easy to integrate...use u=3x+1
oh! hahaha.kk
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wait im stuck
\[\frac{1}{(3x+1)^2}\le\frac{1}{x^2}\] then \[0\le\int\limits_{1}^{\infty}\frac{1}{(3x+1)^2}dx\le\int\limits_{1}^{\infty}\frac{1}{x^2}dx=\cdots\]
what zarkon said, although you know it converges because the degree of the denominator is 2 and the degree of the numerator is 0, and 2 - 0 > 1
wow, satellite73, is that atrick? how does that work?
not sure i would call it a trick, rather it is a fact
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