tjutta
lim x>infinite 1/x^2 S(integral sign) tan^-1 dt
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ash2326
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\[\lim_{x\to \infty} \frac 1 {x^2} \int \tan^{-1} t\ dt\]Is this your question?
tjutta
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yes sorry Im new to this :-)
ash2326
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No need to be sorry :)
ash2326
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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
\[\frac 1 \infty=???\]
ash2326
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@tjutta ???
tjutta
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\[\infty\]
ash2326
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nope
\[\frac 1 0= \infty\]
so
\[\frac 1 \infty=???\]
tjutta
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0
ash2326
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good :D
\[\int \tan^{-1} t\times 0=???\]
tjutta
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SIN/COS ? :-)
ash2326
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if you multiply 0 by tan t then what you'd get?
tjutta
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0
ash2326
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so it's 0
tjutta
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what ? the answer is 0 :-) Ok it makes since if you put it that way but for me whos math retarded its difficult:-)
ash2326
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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
ash2326
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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?
tjutta
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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?
ash2326
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okay, did you try this?
tjutta
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I dont know what to do :-) should I start by divide ?
ash2326
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What's the expansion of (x-1)^2?
tjutta
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(x-1)(x+1) ?
ash2326
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What would you get if you multiply these?
tjutta
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ok I only have 2 ?
ash2326
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\[(x-1)\times (x-1)=???\]
tjutta
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x^2 - 2x + 1
ash2326
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Good:)
Numerator we have x^2, so add and subtract (-2x+1)
Could you rewrite the integral using this?
tjutta
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\[\int\limits_{}^{} \frac{ x }{-2x+1 } \] Im I on the right way? I know Im not very good at this at all :-)
ash2326
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I'm sorry, i thought it's x^2 in numerator
Sorry. Let me think about this