integrate x/lnx

- dan815

integrate x/lnx

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- anonymous

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

- anonymous

you will not find a nice closed form for this one

- dan815

ah

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## More answers

- dan815

is it one of those repeating integrating ones

- anonymous

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

- anonymous

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 \(\frac{x}{\ln(x)}\)

- anonymous

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.

- dan815

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

- dan815

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

- anonymous

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.

- inkyvoyd

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

- inkyvoyd

imo you'd best be off solving it numerically

- dan815

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

- anonymous

Yup

- 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

- anonymous

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

- 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

- dan815

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

- 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

- dan815

but how come*

- anonymous

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

- 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

- dan815

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

- anonymous

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.

- dan815

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

- dan815

where z = ysinx^2

- anonymous

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

- dan815

oh nvm i see what u mean the diagram is wrong

- dan815

so the bottom part shud be the shaded region right

- dan815

|dw:1363061109806:dw|

- anonymous

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

- dan815

yeah

- dan815

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

- anonymous

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

- dan815

0 to 4 and
0 to root x

- dan815

ok got it

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