## tjutta lim x>infinite 1/x^2 S(integral sign) tan^-1 dt one year ago one year ago

1. ash2326

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

2. tjutta

yes sorry Im new to this :-)

3. ash2326

No need to be sorry :)

4. 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 $\frac 1 \infty=???$

5. ash2326

@tjutta ???

6. tjutta

$\infty$

7. ash2326

nope $\frac 1 0= \infty$ so $\frac 1 \infty=???$

8. tjutta

0

9. ash2326

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

10. tjutta

SIN/COS ? :-)

11. ash2326

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

12. tjutta

0

13. ash2326

so it's 0

14. tjutta

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

15. 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

16. 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?

17. tjutta

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?

18. ash2326

okay, did you try this?

19. tjutta

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

20. ash2326

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

21. tjutta

(x-1)(x+1) ?

22. ash2326

What would you get if you multiply these?

23. tjutta

ok I only have 2 ?

24. ash2326

$(x-1)\times (x-1)=???$

25. tjutta

x^2 - 2x + 1

26. ash2326

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

27. tjutta

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

28. ash2326