Question

integral from one to infinity (dx)/(x - 1)^2 ..is it convergent or divergent? my prof tells me it is a twice improper integral..so i don't know what he means by that. when i first attempted the problem i just took the ln of the integral giving me ln(x-1)^2 with b = infinity and lower bound being 1. after evaluating the integral i came up with an answer of infinity. making the equation divergent

Answer 1

convergent bcoz integral will come out to b 1 (finite)

Answer 2

how though?

Answer 3

the lower bound is also not defined, so it has two undefined bounds, hence it is twice improper

Answer 4

also\[\int\frac{dx}{(x-1)^2}\ne \ln(1-x)^2\]

Answer 5

are there certain steps that i must take just because it is twice improper or do i just go about solving the integral normally?

Answer 6

you integrate normally, but you will have to take the limits for both bounds\[\int_a^bf(x)dx=\lim_{n\to a}\lim_{m\to b}\int_n^m\]

Answer 7

\[\int_a^bf(x)dx=\lim_{n\to a}\lim_{m\to b}\int_n^mf(x)dx\]

Answer 8

i took u = x - 1 and my du = dx. so..\[\int\limits_{?}^{?}du/u^2\] then bringing up the u^2 would make it u^-2 du giving me \[-u ^{-1}\]..plugging in u = x-1 would give me -(x-1)^-1 giving me -1/x-1 as an answer to the integral right? if so, i don't understand what you mean by taking limits for both bounds. what would my bounds be for infinity and 1?