OpenStudy (anonymous):

converge or diverge: from 1 to infinty: 1/(3x+1)^2dx

5 years ago
OpenStudy (anonymous):

@eliassaab medal? lol

5 years ago
OpenStudy (zarkon):

convergent

5 years ago
OpenStudy (anonymous):

please show how to do it

5 years ago
OpenStudy (zarkon):

this one is easy to integrate...use u=3x+1

5 years ago
OpenStudy (anonymous):

oh! hahaha.kk

5 years ago
OpenStudy (anonymous):

wait im stuck

5 years ago
OpenStudy (zarkon):

\[\frac{1}{(3x+1)^2}\le\frac{1}{x^2}\] then \[0\le\int\limits_{1}^{\infty}\frac{1}{(3x+1)^2}dx\le\int\limits_{1}^{\infty}\frac{1}{x^2}dx=\cdots\]

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

wow, satellite73, is that atrick? how does that work?

5 years ago
OpenStudy (anonymous):

not sure i would call it a trick, rather it is a fact

5 years ago
OpenStudy (anonymous):

|dw:1334808862422:dw|

5 years ago
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