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lim x>infinite 1/x^2 S(integral sign) tan^-1 dt

Mathematics
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\[\lim_{x\to \infty} \frac 1 {x^2} \int \tan^{-1} t\ dt\]Is this your question?
yes sorry Im new to this :-)
No need to be sorry :)

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Other answers:

The integral doesn't involve x, so it can be taken out. \[\int \tan^{-1} t\ dt\times \lim_{x\to \infty} \frac 1 {x^2} \] tell me what's \[\frac 1 \infty=???\]
\[\infty\]
nope \[\frac 1 0= \infty\] so \[\frac 1 \infty=???\]
0
good :D \[\int \tan^{-1} t\times 0=???\]
SIN/COS ? :-)
if you multiply 0 by tan t then what you'd get?
0
so it's 0
what ? the answer is 0 :-) Ok it makes since if you put it that way but for me whos math retarded its difficult:-)
You know the integral is a variable of t, if you integrate it and multiply by zero. You'd get zero. So it's not necessary to integrate. It's 0
I doubt that this is so easy. I think you have missed some portion of the question. Maybe limits on the integral. Could you check the question again?
its just as you put it so I think this is right but I have one more :-) its \[\int\limits_{}^{} \frac{ (x^3)\ }{(x-1)^2(x+2) }\] Can you help me with that?
okay, did you try this?
I dont know what to do :-) should I start by divide ?
What's the expansion of (x-1)^2?
(x-1)(x+1) ?
What would you get if you multiply these?
ok I only have 2 ?
\[(x-1)\times (x-1)=???\]
x^2 - 2x + 1
Good:) Numerator we have x^2, so add and subtract (-2x+1) Could you rewrite the integral using this?
\[\int\limits_{}^{} \frac{ x }{-2x+1 } \] Im I on the right way? I know Im not very good at this at all :-)
I'm sorry, i thought it's x^2 in numerator Sorry. Let me think about this

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