what is the integral of this??
970181
1) This has no elementary solution\n2) This one, however, does have a solution, with the substitution e^x = tanu.  You are at the answer arctan(e^x)\n\nSteve
748051
The primitive of your integral is:\n\ns*(e^(st))*(1/w)*atan(s/w)\n\nYou have to do a trigonometric substitution to get there. Do not always trust the weird answers you get with complex calculators.
173783
I believe that the integral of 1/x is equal to the natural log of x + C.  (ln x + C).  Usually you take the antiderivative of the function but there are exceptions to the rule, like 1/x.
800132
how does this fit into mathematics?
772619
integral(x,-1,2) - 2 * integral(|x|,-1,2)\nintegral(x,-1,2) - 2 * ( integral(-x,-1,0) + integral(x,0,2))\n\n3/2 - 2( 1/2 + 2)\n3/2 - 3\n-3/2
147855
Try http://integrals.wolfram.com/
1143501
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
832910
yes the integral of the derivative of f(x)=x^2 is x^2.\n\nand just to check:\nf(x)=x^2\nf'(x)=2x\nF(x)=2*(1/2)x^(1+1)=x^2\n\nwhether you graph it or not makes no difference.
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.
234468
what kinda question is "THIS"?
480262
the integration for any intgral from (a) to (-a) is always equal zero.\nso the intgral for this is zero\nbut if there is no interval the integration is very complicated
1092533
The previous person almost got there.\nThe indefinite integral is (1+x)[log(1+x) - 1] + C\n\nSubstitute x=1 to get 2[log(2) - 1] + C\nSubstitute x=0 to get  -1 + C.\n\nThe definite integral is \n2[log(2) - 1]  - [-1]\n= 2log(2) -1\n=log(4) - 1\n=log(4) - log(e)\n=log(4/e)
957499
The average value formula in the interval (a,b) is\n\nAV = 1/(b-a) * DEFINT f(x)  {evaluated from a to b}\nAV = 1/(b-a) * {INT f(4) - INT f(-4)} \n\nThe integral will be 1/2(x^4 - 5x^2).  If you evaluate this at x=4 and then at x=-4 and do the math it will be 0.\n\nSo the answer is 0.
403275
Look, to count this integral, you must make a subsitution:\n\nu = 1+5t\ndu = 5 dt\ndt = du/5\n\nint 8/(1+5t) dt =\n= int 8/u du/5 =\n= 8/5 * int (1/u) du =\n= 8/5 * ln u + C =\n= 8/5 * ln(1+5t) + C
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
402001
Watch it ….! \n\ndelta\n\n x =t^2  ---> t= sqrt(x)\ndx = 2t dt .............(not dx = dt /2t )\n-->\ndt = dx/(2 Sqrt(x)\n\nand this Integral can not be done using Analytical Method; Also, the (method of Integration by Part) can not apply in such integral because either of the cos(t) and sqrt(t) have infinitive derivatives.
94569
Your integral needs either limits (from 1 to X, for example) or a constant added to it. With limits, the apparent contradiction vanishes. With an arbitrary constant, it can be accounted for.