## dan815 2 years ago integrate x/lnx

1. Xavier

With ln(x) in the denominator like that that's not one of our preatty integrals.

2. satellite73

you will not find a nice closed form for this one

3. dan815

ah

4. dan815

is it one of those repeating integrating ones

5. satellite73

you mean like if you integrate by parts will you go around in a circle? i don't think so

6. satellite73

i think you will not be able to solve in in the sense that you will not be able to find some nice function $$F$$ whose derivative is $$\frac{x}{\ln(x)}$$

7. Xavier

It's the type where our general methods of integrating get us nowhere. We have to leave it in the form of an integral and work with it that way.

8. dan815

this is what im trying to evaultate |dw:1363059332628:dw|

9. dan815

is there another method to equate integral like i know this isnt conservative so

10. Xavier

That's completly different. If you are integrating with repect to y first then you don't need to worry about the expressions in x -- they are essentially constants.

11. inkyvoyd

Wolfram says it's not expressible in terms of elementary functions - http://www.wolframalpha.com/input/?i=integrate+x%2Fln+x

12. inkyvoyd

imo you'd best be off solving it numerically

13. dan815

oh hold on i think i got it xD i forgot the lnx cancels if i do dy first

14. Xavier

Yup

15. dan815

xavier u got some time to help ?? im working on one of my assigments due in about 6 hours, gonna need help when i get stuck

16. Xavier

I'm here on and off. I'll try to help. Sure other people will jump onboard if I take too long

17. dan815

just looking thru the questions so far i think im gonna just have some trouble in the volume integral questions ill tell ya when i get to them

18. dan815

look at this one its about setting the right bounds |dw:1363060518856:dw|

19. dan815

i get why for dx its y^2 to 4 but how much for dy just doing 0 to 2 instead is enough how come it cant be be rootX to 4

20. dan815

but how come*

21. Xavier

What is correct there? The diagram or the bounds on your integrals?

22. dan815

the whole thing is correct i just need to covert to an another double integration where im doing it in reverse order dy dx

23. dan815

but before i do that i wanna know why they set up bounds like that

24. Xavier

Either the bounds are wrong or the shaded area is incorrect. Having $y^2 \le x \le 4$ is the larger of the two areas there.

25. dan815

the shaded region is what im integrating over its in the bounds the function is just like a force equation

26. dan815

where z = ysinx^2

27. Xavier

Then your bounds on x should be $0 \le x \le y^2$

28. dan815

oh nvm i see what u mean the diagram is wrong

29. dan815

so the bottom part shud be the shaded region right

30. dan815

|dw:1363061109806:dw|

31. Xavier

Ah ok so you know that the curve your have is y^2=x or y=sqrt(x). So y is bounded above by sqrt(x) and below by 0. And x ranges from 0 to 4

32. dan815

yeah

33. dan815

so if i wanted to switch order of integration to dy dx then

34. Xavier

Just use the bounds on y on the inner integral. x bounds go on outer integral $0 \le y \le \sqrt{x}\ ; 0 \le x \le 4$

35. dan815

0 to 4 and 0 to root x

36. dan815

ok got it