what is the proof for the antiderivative of ln(x)/2x?
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.
883916
Let log_2(x) denote the log of x with base 2.\n\nUse the identity that log_2(x) = ln(x) / ln(2). This identity holds because:\n\nRecall that, by definition, 2^(log_2(x)) = x.\n\nln(x) = ln(2^(log_2(x))) = (log_2(x)) (ln(2)), so dividing through by ln(2) shows the result.
549061
1/x^2 can be rexpressed as x^(-2). As you might know, the formula for a derivative of x^n is nx^(n-1). The antiderivative of x^n is (x^(n+1))/n+1. In this case, it would be x^-1/-1, or -x^-1. To check this, we take the derivative, which gives us (-1)(-1)x^-2, or x^-2. The answer checks out. \n\nSo, in summary, the antiderivative of x^-2 is -x^-1, with the remainder C.
275631
One way to see this is to write x^(1-x) as exp((1-x)ln x) and use the chain rule. Only one person has the correct answer, art_colls. The power rule *does not apply* since the exponent is a function also!
823238
I wont give you the entire proof, just give you the key arguments.\n\nknowing the joint distibution of (X, Y) allows to compute the marginal densities of X and Y.\nIndeed, let f_X be the marginal density of X, then:\n\nf_X(x) = &#92;int f(x, y) d_y\n\nf_Y(y) = &#92;int f(x, y) d_x\n\nNow, you might compute E(Y| X = x) = &#92;int y f_{Y|x} (y) d_y\n\nwhere f_{Y| x} is the conditionnal density of Y knowing the event X = x\n\nrecall that f_{Y| x}  = f(x, y) / f_X(x), and you just have to compute the resulting integrals\n\nregards,\n\njfp
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works
217134
Hi there,\nThis is a linear differential equation and can be solved using a certain list of steps (check any calculus text book). To avoid finding an integrating factor etc, we note that by dividing by x^2 we get:\nf'(x)/x - f(x)/ x^2 = 1/x\nThe left side is the same as (f(x)/x)' [product rule].\nThus we have, (f(x)/x)' = 1/x\nIntegrating gives f(x)/x = ln|x|+C\nSo, f(x)=x ln|x|+Cx\nSubstituting f(-1)=1 gives C=-1\nThus, f(e^-1)=(e^-1) ln|e^-1| - e^-1=-2e^-1.\n\nCheerios,\nMike
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.
1022433
To perhaps elaborate on nardneran's answer:\n\nIn this problem you use the substitution rule. Let u=ln x. Then du=1/x dx and dx=x du. Replacing ln x with u and dx with x du gives ∫u² du (the x's in your equation cancel). You can integrate that to get u³/3 + C and back-substituting gives (ln x)³/3 + C.
195729
you dirty dirty man... making me derive... *dusts off brain* alright, give this a shot...\n\nstarting with the inner terms....\n\nderivative of x... 1\n\nderivative of 4sin(x)... 4cos(x)\n\nderivative of 2... 0\n\nderivative of the larger term 1/(2(4sin(x)+2)^(1/2))\n{term needs to be taken to the bottom of the fraction as the exponent is negative}\n\nCompiled we get... 4cos(x)/(2(4sin(x)+2)^(1/2))\n\ninserting the 0... we get... 4/(2(2)^(1/2))... or 2/2^(1/2)\n\nrationalize... multiply by 2^(1/2)/2^(1/2) and we end up with....\n\n2^(1/2)
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
347784
Taking natural logarithm to both sides\nLn(y)= Ln(x) * Ln(x) = (Ln(x))^2\nDifferentiating\n(1/y )dy/dx = (2/x) * (Ln(x))\ndy/dx = (2y/x) * (Ln(x))\nat the point (e,e),\ndy/dx = (2e /e) * Ln(e)\nso the slope = 2
316594
The anti-derivative of x^(2)*e^(x)^(3) is (1/3)*e^(x)^(3)\n\nYou can always check by taking the derivative of your answer.
746085
taking log on both sides\nlny=(x-5)ln2\n1/y(dy/dx)=ln2\ndy/dx=yln2=2^(x-5)ln2
789574
from the ⌠udv = uv - ⌠vdu, let u=(lnx)^2 and dv=x^4dx\nhence du=2dx/x and v=x^5/5\nsubstituing values we have:\nx^4 (lnx)^2 dx = (lnx)^2(x^5/5) - ⌠(x^5/5)(2dx/x)\n                      = (lnx)^2(x^5/5) - ⌠(2/5)x^4dx\n                      = (lnx)^2(x^5/5) - x^5/10\nthen using the limits; this yields to 1.796442507
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|