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|dw:1350067985984:dw|

I have the solution, looking for help to understand the process

lol .. "Improper Integral" is not quite the term you describe it.

break x^4 in (x^2)^2 then use substitution

|dw:1350068067914:dw|

yes

are you familiar with last three special integrals formula

three special integral formula? not sure which one you are referring to

did you use double substitution

let x^2 = u sub ... this would end up into inverse typerbolic function.

|dw:1350068248916:dw|these two?

well ... you can try that. but I'll give you a shortcut.

ok I will love to see the shortcut

look at the the inverse hyperbolic of sine
http://en.wikipedia.org/wiki/Inverse_hyperbolic_function

@experimentX me to please

|dw:1350068353134:dw|

@experimentX nice one

I see it is the first one

|dw:1350068751083:dw| solution manual has this as a second step

hmm ... it really was improper intgral!!

Is this because of the square root function's domain being continuous on [a, infinity)?

this doesn't converge.

ok I am following you so far

|dw:1350068915623:dw|

|dw:1350069081160:dw|

In the definition of Improper integrals over infinite intervals
I have 3 definitions

It is improper integral because of that infinity thing.

|dw:1350069479305:dw|
provided the limit exists

I should mention that if f is continuous on [a,infinity), then
for the above first definition

|dw:1350069732672:dw| this is called improper integral of First Kind.

|dw:1350069792525:dw|

|dw:1350069861584:dw|

the second kind appears when you have singularity in domain.|dw:1350070237179:dw|

|dw:1350070363244:dw|

|dw:1350070434570:dw|

try watching this
http://www.youtube.com/watch?v=KhwQKE_tld0

ok thanks, I can use all the help

Thanks, I enjoyed that lecture.