can any body solv it integration of e to the power x to the power x { e^x^x}???
488973
If you mean this (e^x)^x, then that is e^(x^2).  There is no closed form expression for the antiderivative of this with respect to x.\n\nHowever if you mean e^(x^x), then I don't know.\n\nUPDATE: Hey discoganya, I'm getting d/dx(x^x) to be \n(x^x)(ln x + 1), not x^x.  Now (d/dx)e^x = e^x.\n\nUPDATE (again): I think that part of the problem with that, discoganya, is that you are assuming that x is a whole number.\n\nThis is how I got my dervative.\n\n(d/dx)(x^x) = (d/dx)(exp(xln x))\n= exp(xln x)(1*ln x + x(1/x))\n= (x^x)(ln x + 1), for x > 0\n\nI have run this through Mathematica, and it confirms it.\n\nI believe the only function such that f'(x) = f(x) is f(x)=  a*exp(x), where a is an arbitrary constant.\n\nBy the way, Mathematica also could not find the integral of e^{x^x), honey, so it also probably does not exist in a closed form.
759209
This is a non-elementary integral in x. However, from the way you phrased the problem, I would guess that you want this to compute a double integral: first with repspect to x, then with respect to y. The best way of doing *that* would be to switch the order of integration. Then the integral with respect to y will be xe^(y/x). Be sure to change the limits of integration so that you are integrating over the same region in the xy-plane!
935326
You have got to use expansion series!\n\nX->0+ [e-(1+x)^(1/x)]/x =e/2 \nlet x=0+h\n=>as x->0, h->o\n\nh->0 [e-(1+h)^(1/h)]/h\n\nlet y=(1+h)^(1/h)\n=>log y=(1/h)log(1+h)\n=>y=e^[(1/h)log(1+h)]\n\nexpansion of log(1+x)=x - (x^2)/2..................\n=>e^[(1/h)log(1+h)]= e^[(1/h)[h - (h^2)/2!............]\n\n=>e^[(1/h)[-h - (h^2)/2!............]=e^[1-h/2!........]\n\ntherefore,\n\nlet y =lim h->0 [e-(1+h)^(1/h)]/h=lim h->0 {[e- e^(1-h/2!...)]/h}\n\nexpansion of e^x=1+x+(x^2)/2!.................\n\n=>lim h->0 {[e- e^(1-h/2...)]/h}=lim h->0 {-e[e^(-h/2.....) -1]}/h\n\nexpansion of e^x=1+x+(x^2)/2!.................\n=>lim h->0{-e[1+(-h/2-(h^2)/3.....)+ ........-1]/h}\n\n=>lim h->0{-e[(-h/2-(h^2)/2...)+ [(-h/2-(h^2)/2...)^2]/2!+.....]}/h\n\n=>lim h->0{-e(-1/2 -..(terms containing highter powers of h))}\n\n=>e/2\n\n\nP.S.  By the way L' Hospitals rule would fail here as limit is not of form 0/0 or ~/~. It is of form a/0 where a is constant.
1115014
It's e^x^2 + C,  but i don't know what other calculus it could have formed.  Try searching for the History of Integrations?
787465
Write out the exponentials and use the series for e^x. Other than that, do your own homework.
754435
I | AUG is not correct.  You cannot differentiate power functions by the exponent rule.  (Even doing that, she made an error.)  The implicit differentiation done by Greg G is correct, or you can use the functional power rule:\n\nd(ƒ^g)/dx = (ƒ^g)*[(dƒ/dx)*(g/ƒ) + (dg/dx)*lnƒ]
934968
Okay, assuming you mean:\n∫e^(2x)/√(1-e^(2x)) dx\nLet u=1-e^(2x), du=-2e^(2x) dx, dx = -1/2 e^(-2x)\n-1/2 ∫1/√u du\n-1/2 ∫u^(-1/2) du\n-u^(1/2) + C\n-√(1-e^(2x)) +C
756165
In general e^(i*x)=cos(x)+i*sin(x). Plug in x=pi to get the equation you wanted.\n\nOf course, this just pushes off the result that needs to be demonstrated to the formula for e^(i*x). There are a couple of ways to show this. The first simply plugs i*x in for x in the power series for e^x. The powers of i are regular and when you break the series into even and odd terms, you get the series for cos(x) and sin(x).\n\nAlternatively, realize that e^(ix) is a solution of the differential equation\ndy/dx=i*y\nand so is cos(x)+i*sin(x). Both also satisfy y(0)=1. By uniqueness results for solutions of differential equations, the two must be equal.
1143501
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
729281
The  exponent becomes a coefficient\n-1+2x lne\nln e=1\n-1+2x
808989
Let t=sinx\n=> dt = cosx dx\n\nx=pi/2 => t=1\nx=0 => t=0\n\nI=integrate(e^t dt) from 0 to 1\n\n=e^t from 0 to 1\n\n=> I= e^1 - e^0 = e-1\n\nAnswer is I=e-1
408775
1/(1+e^x)=(1+e^x -e^x)/(1+e^x)=1-  e^x/(1+e^x)=\n1- (1+e^x)'/(1+e^x)\n& u'll integrate
171290
I think you mean "integral" of x^(n-1)* exp(-(log)^2)/2 and I THINK that the e and the log should cancel so you would have X^(n-1) * -X^2/2 = x^n-1+2 /2 so the integral of that would be... X^(n+1)/(2n+4)
278447
answer No 1 is bullshit. Answer No 2, I couldn't even understand what he/she wants to say.\nBut answer 3 is correct:\nint(e^(x-y)=int(e^-y e^x) = e^-y int(e^x) = e^-y e^x +c=\ne^(x-y) +c
916864
Ord(e^(4x)-1 -4x) ~ Ord(e^(4x) > Ord(x^2)\nThen:\n\nlim(e^(4x)-1 -4x)/x^2 {x--> oo} = oo\n\nRemember\n\nln x < x^a < e^x < x^x
276671
(e^(cose^e+sin^e)/2)^pi\nhahaha ..just kidding
1082973
I answered in your other post but I'll re-post the answer here:\n\nYou elevate the number to the power 2 like this:\n\nx^2\n\nYou should still do the normal product but it is easier to thing you et the square of a certain number and most calculators already have the elevate to the second power function integrated\n\nCynthia