anonymous
  • anonymous
lim x>infinite 1/x^2 S(integral sign) tan^-1 dt
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
schrodinger
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

ash2326
  • ash2326
\[\lim_{x\to \infty} \frac 1 {x^2} \int \tan^{-1} t\ dt\]Is this your question?
anonymous
  • anonymous
yes sorry Im new to this :-)
ash2326
  • ash2326
No need to be sorry :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ash2326
  • ash2326
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
  • ash2326
@tjutta ???
anonymous
  • anonymous
\[\infty\]
ash2326
  • ash2326
nope \[\frac 1 0= \infty\] so \[\frac 1 \infty=???\]
anonymous
  • anonymous
0
ash2326
  • ash2326
good :D \[\int \tan^{-1} t\times 0=???\]
anonymous
  • anonymous
SIN/COS ? :-)
ash2326
  • ash2326
if you multiply 0 by tan t then what you'd get?
anonymous
  • anonymous
0
ash2326
  • ash2326
so it's 0
anonymous
  • anonymous
what ? the answer is 0 :-) Ok it makes since if you put it that way but for me whos math retarded its difficult:-)
ash2326
  • ash2326
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
  • ash2326
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?
anonymous
  • anonymous
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
  • ash2326
okay, did you try this?
anonymous
  • anonymous
I dont know what to do :-) should I start by divide ?
ash2326
  • ash2326
What's the expansion of (x-1)^2?
anonymous
  • anonymous
(x-1)(x+1) ?
ash2326
  • ash2326
What would you get if you multiply these?
anonymous
  • anonymous
ok I only have 2 ?
ash2326
  • ash2326
\[(x-1)\times (x-1)=???\]
anonymous
  • anonymous
x^2 - 2x + 1
ash2326
  • ash2326
Good:) Numerator we have x^2, so add and subtract (-2x+1) Could you rewrite the integral using this?
anonymous
  • anonymous
\[\int\limits_{}^{} \frac{ x }{-2x+1 } \] Im I on the right way? I know Im not very good at this at all :-)
ash2326
  • ash2326
I'm sorry, i thought it's x^2 in numerator Sorry. Let me think about this

Looking for something else?

Not the answer you are looking for? Search for more explanations.