Question

show that integral x / (x^2 + 2 ) ^2 converges on [ 0,oo), use integral comparison

Answer 1

probably can compare it to
\[\int_0^{\infty}\frac{x}{x^4}dx\] since it is smaller

Answer 2

no doesnt work, that diverges

Answer 3

at least for the infinity part

Answer 4

integral 1/ x^3 diverges on [ 0, oo)

Answer 5

this is not an improper integral on the left, because it is finite at 0.

Answer 6

in fact it is 0 at 0

Answer 7

you mean the integrand is zero at zero ,
integral f(x) , f(0) = 0

Answer 8

right. the integrand

Answer 9

x/ (x^2 + 2 ) ^2 is zero when x = 0
but the comparison wont work

Answer 10

sorry that is not what i mean

Answer 11

i mean to say
\[\int_0^1\frac{x}{(x^2+2)^2}dx\] is finite

Answer 12

and
\[\int_1^{\infty}\frac{x}{(x^2+2)^2}dx\] converges by comparison test

Answer 13

that other stuff i wrote was nonsense, sorry

Answer 14

in fact i think you can just go ahead and compute this integral if you like, because you can use a simple u - sub to get the anti derivative

Answer 15

it is
\[-\frac{1}{2(x^2+2)}\]
at 0 you get
\[\frac{-1}{4}\] and as r goes to infinity you get 0

Answer 16

so integral is in fact just
\[\frac{1}{4}\]

Answer 17

so if it wasnt easily integrable, you would have to split the integral