suppose the integral of..?
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.
833486
int1/x^2+int1/x^5\nx^-1/-1 +x^-4/-4+C\n-1/x-1/4x^4+C
414104
Using an integral,\n\nWords = S 3t(3t^2 + 1)^(-1/2) dt\n\nlet u = 3t^2 + 1\nthen du = 6t dt\n\n= (1/2) S u^(-1/2) du\n= (1/2)(2u^(1/2))\n= u^(1/2)\n= (3t^2 + 1)^(1/2)\n\nEvaluating from 1 to 4,\n= (3(4)^2 + 1)^(1/2) - (3(1)^2 + 1)^(1/2)\n= 49^(1/2) - 4^(1/2)\n= 7 - 2\n= 5\n\nAnswer A.
496092
let x equal (tant)^2=>dx=(sect)^2dt. then simply the integrand, u will have something like integral of (cost)(sint)^2 dt. let u be sint. then u know it :D
482644
Let us denote the integral sign by I.\nWe will use substitution method.\nLet x = 2*tan(t)\nThen dx = 2*sec^2(t)dt\nThus, I[1/(x^2+4)^2]dx\n= I[1/{4*tan^2(t)+4}^2]*2*sec^2(t)dt\n= (1/8)*I[1/{tan^2(t)+1}^2]*sec^2(t)dt\n= (1/8)*I[1/{sec^2(t)}^2]*sec^2(t)dt  [since tan^2(t) + 1 =sec^2(t)]\n= (1/8)*I[1/{sec^2(t)}]dt\n= (1/8)*I[cos^2(t)]dt\n= (1/16)*I[2*cos^2(t)]dt\n= (1/16)*I[1+cos(2t)]dt [since 2*cos^2(t) = 1 + cos(2t)]\n= (1/16)*I[dt] + (1/16)*I[cos(2t)]dt\n= (1/16)*t + (1/16)*[{sin(2t)}/2] + C where C = arbitrary constant of integration\n= t/16 + {sin(2t)}/32 + C\n\nNow sin(2t) \n= {2*tan(t)}/{1+tan^2(t)}\n= 4x/(4+x^2) [since x = 2*tan(t)]\n\nAlso t = tan-1(x/2) [tan-1(x/2) means tan inverse (x/2)]\n\nHence the given integral equals to\n{tan-1(x/2)}/16 + x/{8(4+x^2)} + C
1090386
What?\nIf u=x^x\ndu/u=(1+ln(x))dx\n\nInteral  exp(x^x)dx=integral exp(u)du\n\nAnalytical solution for this indefinite integral is dificult.
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
614800
It converges to one.\nNote that |x| = x if x > 0 and |x| = -x if x < 0.\nBreak the integral into two pieces.\nint from -infinity to 0 of -xe^(-x^2) dx \n+ int from 0 to infinity of xe^(-x^2) dx\nUse substitution letting u = -x^2\nint from -infinity to 0 of (1/2)e^u du\n+ int from 0 to -infinity of (1/2)e^u du\n= (1/2)e^u from -infinity to 0 -(1/2) e^u from -infinity to 0\n= (1/2)[e^0 - 0] - (1/2)[0 - e^0]\n= (1/2)(1) - (1/2)(-1)\n= 1\nI don't have this typed quite correctly.  You have to replace -infinity with b and then take the limit as b goes to -infinity.
83265
That's an elliptic integral.  Check out the link.
1143501
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
832686
Take the integral of this function, like this:\n\n∫ (10 + x²) / x³ =\n\n∫ (10 / x³) + ∫ (x² / x³) =\n\n10 ∫ (1 / x³) + ∫ (1 / x) =\n\n10 (-1 / 2x²) + ln(x) + C = \n\n-5 / x² + ln(x) + C\n\n~ ♥ ~
297440
Are you looking for indefinite  integral   of x^(sin(1/x)) ?\n\nMathematica has a web site you can plug-in the integrant and get the result (if it exist in closed form)\n\nBy hand: try to break this power expression using log and exp functions. If  a=ln(b), then b=exp(a) and therefore b=exp(ln(b)).\n\nIn this way your integrant could be written as \n\nexp(ln(integrand))\nand since it is of power form, use the properties of ln and exp functions to break it down and thus reduce it to simpler integrals.
1060628
simple find the integral of pi*(e^(-x^2))^2 from a=0 and b=2 since as x goes to infiniety is goes to zero but it is very close to zero when x=2 you can find this by graphing f(x) \nanswer should be 0.9188 good luck!
117158
the integration >>> increase the power by one and divide by the new power (don't forget the the constant)\n\nx^n -------> x^(n+1)/(n+1)\n\ny = 1/(x^2)\ny = x^-2\nthe integration = x^(-1)/(-1) + c\n= -x^-1 + c\n= -1/x + constant
345069
This can be shown for X as either continuous or discrete.  Let mu = E(X).\n\nVar(Y) = E(Y^2) - E(Y)^2\n= E[(X-mu)^2] - E(X-mu)^2\n= E(X^2 - 2Xmu + mu^2) - (E(X)-mu)^2\n= E(X^2) - 2muE(X) + mu^2 - (mu-mu)^2\n= E(X^2) - 2mu^2 + mu^2 - 0\n= E(X^2) - mu^2\n= E(X^2) - E(X)^2\n= Var(X)\n\nIf you need to be more specific than this, the replace E(X) = int(-inf to inf) xf(x)dx and E(X^2) = int(-inf to inf) x^2f(x)dx.  Everything should come out the same way.\n\nRemember that using substitution in the integral, let x = y + mu, the dx = dy, and the limits of integration would remain -inf to inf.
696867
cos[x]dx=dsin[x] therefore the integral is over e^y in the limits y=sin[Pi/2]=1 to y=Sin[0] = 0; and is equal to 1-Exp[1]
74089
http://integrals.wolfram.com/