does the integral of ln x exist?
155864
yes, it equals x(ln(x)-1)
452294
Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
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)
337080
when you integrate 1/x you get In |x|\n\nso the answer is -1/2 * In |x| + constant term\n\n\nYou can use substitution, but there's not much point, because you'll get 1/u then (which is the same thing). 1/x is a well known integrand. You'd be better just memorising that the integral of 1/x = In(x)
796927
try the link below...... hope it can put an end to ur integral problems
14407
This is an integration-by-parts problem. The answer is:\nx*ln(x)+C\n\nThis is how you calculate it:\n\nIntegral of ln(x)*dx\n\nu= ln(x)\ndu=1/x*dx\ndv=dx\nv=x\n\nSo by parts: uv-integral of vdu\n= ln(x)*x-integral of x*1/x*dx\n=x*ln(x)-integral of 1*dx\n=x*ln(x)-x+c
26362
Part 1:\n\nWe can rewrite y^x as (e^ln(y))^x which is e^(xln(y))\n\n∂f/∂x f(x,y)= ∂f/∂x (x^y) + ∂f/∂x(y^x)\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)e^(xln(y))\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)y^x\n\n\nPart 2: There is no simple function which is the integral of e^(x^2).
900298
We know that F(x) = ∫ln³ x/x dx + C. So we need to find that integral, solve for the constant of integration, and then plug 9 into the resulting formula. First, the integral:\n\n∫ln³ x/x dx\nLet u=ln³ x, v=ln x, du=3 ln² x/x dx, dv = 1/x dx. Integrating by parts:\n∫ln³/x dx = ln^4 x - ∫3 ln³ x/x dx + C\nAdding 3∫ln³ x/x dx to both sides:\n4∫ln³/x dx = ln^4 x + C\nDividing by 4:\n∫ln³/x dx = ln^4 x + C\n\nNow, solving for C:\n\nF(1) = ln^4 1 + C\nF(1) = 0^4 + C\nF(1) = C\nBut F(1) = 0, so C=0.\n\nNow, solving for F(9):\n\nF(9) = ln^4 9 = (ln 9)^4 ≈ 23.30761
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
845874
1.  This is equal to\nk*(lnx)/x, and since \n(lnx)/x --> 0, so does this expression.\n\n2. Sorry, blank on this and 3.\n\n4. (n^k)/2n = (n^(k-1))/2, and I guess as long as \nk > or = 1\nthis will approach 0.\n\nOf course, if k < 1 then it has no limit, or "approaches infinity"
750014
a. The domain is (0, ∞) This is found by noticing that since x is in the denominator, it can not equal zero, and also because ln(x) is only defined for x > 0.\n\nb. This is asking to solve the equation 0 = (lnx)/x. Since x can not equal zero, this is true when lnx = 0. This is easily found by solving the equation e^(lnx) = e^0. So whenever x = 1.\n\nc. I prefer to use the product rule instead of the quotient rule. So we can rewrite this equation as (lnx)(x)^(-1). So f' = gh'+hg', and g = lnx and h = x^(-1). So g' = (1/x) and h' = (-1/x^2). So f' = -(lnx)/(x^2) + (1/x^2) = (1 - lnx)/(x^2).\n\nd. This is increasing whenever f' > 0. So whenever [(1-lnx) > 0 and x^2 > 0] or [(1 - lnx)<0 and x^2 < 0]. Since x^2 is always > 0, because x > 0, we only need to worry about whenever (1 - lnx) > 0. Solving this inequality gives us whenever 1 > lnx. So, whenever x < e this inequality is true. That means that the function is increasing on the interval (0,e).\n\ne. Using our results from the previous question, and knowing that f is decreasing whenever f' < 0, we can see that this is true whenever 1 - lnx < 0 and x^2 > 0. Once again since x^2 > 0 for all x in the domain, we need only concern ourselves with 1 - lnx < 0. Solving this inequality gives us 1 < lnx. This is true whenever x > e. So the function is decreasing on the interval (e,∞).\n\nf. The relative maxima/minima are where f' = 0. Once again, we need only concern ourselves with when 1 - lnx = 0, since x^2 can not equal zero. So whenever 1 = lnx, or whenever x = e. Since we are not asked to classify this as a maxima or a minima, this answers the question.\n\nIf you want to know, it is a maximum and this can be determined using the second derivative.\n\ng. Now we will find the second derivatice, f''. Once again, I like to use the product rule. Since f' = (1-lnx)/x^2 we can say this is (1 - lnx)(x^(-2)). So g = (1 - lnx) and h = x^(-2) and g' = -(1/x) and h' = (-2/x^3). So f'' = (1 - lnx)(-2/x^3) + (-1/x^3) = (2lnx - 3)/(x^3).
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.
147855
Try http://integrals.wolfram.com/
219346
The first link below is a problem discussing improper integrals (which classifies all integrals with an infinite limit of integration). Just scroll to the bottom of the page--the top of the page is a different problem altogether. Basically, the tip is to replace the "troublesome" limit values with a constant (like k) and then evaluate the limit as k approaches that value. In this case, k becomes one of the limits of integration and you evaluate the integral as k approaches infinity.\n\nI recommend The Complete Idiot's Guide to Calculus (follow the second link in the Sources list below) when brushing off calculus rust, and not only because I am the author of the book. :) It's a light-hearted but content-heavy review of calculus, beginning with advanced algebra review and ending with function approximation by infinite series approximation.\n\nAlso, spend some time with "Tutorials for the Calculus Phobe," free Flash-based calculus tutorials on my website, Calculus-Help.com. They're fun but help you conceptualize calculus basics. The third and final link in the sources houses the Phobe tutorials.\n\nBest of luck!
212677
Little reply: No.\n\nP.S. 12.08 AM GMT\nThis gives the Fresnel Integral S(x), more exactly\n\nS(x)=Integral(.5Pi*x^2)\n\n(the integral is from 0 to x), and it is not possible to reduce it in a finite way.\n\nHowever S(x) has an asymptotic expansion for x>>1\n\n.5-(cos(.5Pix^2))/(x*Pi)\n\nwhich gives limit S(infinity)=0.5
920193
e^(xlnx )=x^x\nderivative is  1+ln x