Question

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

Answer 1

\[\lim_{x\to \infty} \frac 1 {x^2} \int \tan^{-1} t\ dt\]Is this your question?

Answer 2

yes sorry Im new to this :-)

Answer 3

No need to be sorry :)

Answer 4

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
\[\frac 1 \infty=???\]

Answer 5

@tjutta ???

Answer 6

\[\infty\]

Answer 7

nope
\[\frac 1 0= \infty\]
so
\[\frac 1 \infty=???\]

Answer 8

0

Answer 9

good :D
\[\int \tan^{-1} t\times 0=???\]

Answer 10

SIN/COS ? :-)

Answer 11

if you multiply 0 by tan t then what you'd get?

Answer 12

0

Answer 13

so it's 0

Answer 14

what ? the answer is 0 :-) Ok it makes since if you put it that way but for me whos math retarded its difficult:-)

Answer 15

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

Answer 16

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?

Answer 17

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?

Answer 18

okay, did you try this?

Answer 19

I dont know what to do :-) should I start by divide ?

Answer 20

What's the expansion of (x-1)^2?

Answer 21

(x-1)(x+1) ?

Answer 22

What would you get if you multiply these?

Answer 23

ok I only have 2 ?

Answer 24

\[(x-1)\times (x-1)=???\]

Answer 25

x^2 - 2x + 1

Answer 26

Good:)
Numerator we have x^2, so add and subtract (-2x+1)
Could you rewrite the integral using this?

Answer 27

\[\int\limits_{}^{} \frac{ x }{-2x+1 } \] Im I on the right way? I know Im not very good at this at all :-)

Answer 28

I'm sorry, i thought it's x^2 in numerator
Sorry. Let me think about this