What is the integral of e to the x-squared dx?
352249
This integral does exist (it can be evaluated for any endpoints you choose). Since it cannot be represented as a finite combination of elementary functions and operations, it is given its own name, the error function. Many integrals like this have their own names and are quite useful in applied mathematics. See http://mathworld.wolfram.com/Erf.html .\nTo say that it does not exist is unrigorous.
749046
well the term you've quoted is a constant...\n\nso (e^2/1+e^2)x would be the answer to that question.\n\nbut i'm guessing you typed it wrong, or are confused.
263227
This integral does not have an indefinite closed form of elementary functions. It can be evaluated numerically by either integrating the series representation of e^(-x^2) or using integral approximation methods like Simpson's rule.\nIt also has an interesting value if evaluated over its entire domain; by changing coordinates to polar coordinates, one can easily show that the integral over the entire real line is equivalent to Sqrt[Pi].\nThis fact is used in statistics to normalize the Gaussian distribution to have area 1 under the curve. The integral of your expression on the interval [0,t] is then known as the error function. See the source for details.
561041
Do you mean E(X)? That would be the mean value of X, assuming we have a normal (Gaussian) distribution.
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?
68076
god man.  what kind of horrible painful calculus class are you taking?\n\nfirst integrate over dx.  the integral of x^y dx is just a power integral, so you have y*x^(y-1).  the integral of y^x dx is just an exponential, so you have y^x/ln y.\n\nnow integrate y*x^(y-1) + y^x/ln y over dy.  and there, my friend, is where i stop because frankly i can't be arsed to do something so brain-curdling.  actually the first term doesn't look so bad, i think if you integrate it by parts it should come out in one step.  don't really want to think about doing the 2nd one.\n\nyou might consider trying some kind of fancy thing like transforming to polar coordinates or something.  might make it easier.\n\nbtw you should know about this site:\nhttp://integrals.wolfram.com/
226022
Let u = cot(x)\ndu = -csc^2(x)dx\n\nThis integral will then simplify to\n∫(-5*e^(u)du)\n     =-5*e^(u)\n     =-5*e^(cot(x))
1051703
collect like terms on one side\nxdx/(1+x^2)=dy/[(1+y^2)*y]\n\n1/2* 2x/(1+x^2)= dy/y-1/2*2y/(1+y^2)\n\nBy integrating each side,\n1/2*ln(1+x^2)\n=ln|y|-1/2*ln(1+y^2)\n\nuse log rules, take square of both sides,\n\nC*y^2 / (1+y^2)=1+x^2\n\ny=sqrt([1+x^2]/[C-1-x^2])
147855
Try http://integrals.wolfram.com/
821024
∫cos(x^2) dx = √(π/2)FresnelC(√(2/π)x)
1106006
Let u = cx+d, then du = c dx and so (1/c) du = dx\nSolve for x\nu = cx + d\nu - d = cx\nx = (u-d)/c\nThe new integral is\n[a*((u-d)/c) + b]/u (1/c) du\n= (1/c)[(a/c)u - ad/c +b]/u du\n= (1/c)[(a/c) - ((ad/c) + b)(1/u)] du\n= (1/c)[(a/c)u - ((ad/c) + b)ln(u)] + K, where K is a constant\nReplace u with cx+d for your final answer.
729281
The  exponent becomes a coefficient\n-1+2x lne\nln e=1\n-1+2x
796927
try the link below...... hope it can put an end to ur integral problems
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
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
408576
its a logarithmic integral.\ndifferentiation of log x is 1/x\ntherefore integration of 1/a-x is -log(a-x)+c\nnegative sign because coefficient of x is -1
957759
a. 0 = 7-x^2, x^2=7, x = +/-sqrt(7)\nArea = Integral[-sqrt(7), sqrt(7)] (7-x^2)dx = 7x - (1/3)x^3 |-sqrt(7), sqrt(7) = 2[7sqrt(7) - (1/3)sqrt(7)^3] ~= 24.69 u^2\n\nb. 0 = x^4-8, x^4 = 8, x = +/-8^(1/4)\nArea = (1/5)x^5-8x |-8^(1/4), 8^(1/4) = 2[(1/5)8^(5/4)-8(8)^(1/4)] ~absv= 21.53 u^2