How do you integrate sqrt (1-4x^2)/x?
365246
mathematician's approach is the only correct approach in this thread. The square root function does not distribute over addition/subtraction. Consider Sqrt[1 + 1] vs. Sqrt[1] + Sqrt[1].\nIf you're more of a geometer than an algebraist, you will get the same answer from considering the right triangle with leg of length Sqrt[1-4x^2], implying hypotenuse of length 1 and the other leg of length 2x. Your expression is then 2tan(t) where t is the angle opposite the leg of length Sqrt[1-4x^2]. Also, x is obviously (1/2)cos(t), so dx = -(1/2)sin(t) dt.
939593
(x^2-x^(3/2))/2x^2 = 1/2-sqrt(x)/2\n        =1/2(1-x^(-1/2))\n\nd(1/2(1-x^(-1/2)))/dx\n=1/2(-(-1/2x^(-3/2))\n= 1/4x^(-3/2)\n\ni hope that this helps
734906
the integral of [cos 2x dx] is [1/2 sin 2x + c]
800980
1. Move the constant to the right side of the equation:\n4x^2 + 2x = 1\n\n2. Divide out the x^2 coefficient:\nx^2 + 1/2x = 1/4\n\n3. Divide the coefficient of x (in this case, 1/16) by 2 and square it, then add it to both sides:\nx^2 + 1/2x + 1/16 = 5/16\n\n4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:\n(x + 1/4)^2 = 5/16\n\n5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:\nx + 1/4 = +/-sqrt(5/16)\n\n6. Move the constant on the left to the right and simplify:\nx = -1/4 +/-sqrt(5/16) = -1/4 +/-(sqrt 5)/4 = (-1 +/- sqrt 5)/4
427075
If you took the 3 from the denominator, you'd have\n\n1/3 int{(x^2 + 2x - 2) / (sqrt(x)}\n\nsince you can move constants in and out of an integral with no problems.  If you got rid of the radical, you'd then have\n\n=1/3 int{(x^2 + 2x - 2) / x^(1/2)}\n=1/3 int{x^2/x^(1/2) + 2x/x^(1/2) - 2/ x^(1/2)}\n=1/3 int{x^(3/2) + 2x^(1/2) - 2x^(-1/2)}\n=1/3[(2/5)x^(5/2) + 2(2/3)x^(3/2) - 2(2)x^(1/2)]\n=(2/15)x^(5/2) + (4/9)x^(3/2) - (4/3)x^(1/2)\n\nEvaluating from 1 to 8 gives you\n\n=[(2/15) x (8)^(5/2) + (4/9) x (8)^(3/2) - (4/3) x (8)^(1/2)]\n-[(2/15) x (1)^(5/2) + (4/9) x (1)^(3/2) - (4/3) x (1)^(1/2)]\n\n= [256sqrt(2)/15 + 64sqrt(2)/9 - 8sqrt(2)/3] - [(2/15) + (4/9) - (4/3)]\n\n= 968sqrt(2)/45 - 34/45\n\n= (968sqrt(2)-34) / 45
786513
put x^2=t\n2xdx=dt\nxdx=1/2dt\nthe integral now is\n(1/2dt/1+t^2\n(1/2)tan^-1t+C\n=(1/2)tan^-1(x^2)+C
955693
i don't understand the way you\nhave written down the function\n\nsay that you want to integrate\n2x/(2x-1)\nput u=2x-1,>>>x=(u+1)/2,\ndx=du/2,\nint{2x/(2x-1)}dx\n=int{(2*(u+1)/2)/u}*du/2\n=1/2*int{(u+1)/u}du\n=1/2*int{1+1/u}du\n=1/2[u+lnu]+C\nbut u= 2x-1\n=1/2[2x+ln(2x-1)-1]+C\nwhere C is a constant\n\ni hope that this helps
888834
Integrate from -π/2 to π/2 and double it
830100
'x= rcost\ny=rsint\n\nand r= sqrt(x^2+y^2)\nt=arctan(y/x)\nso\nif x=2, and y=2, then r= sqrqt(4+4)=sqrt(8)\nand t=arctan(2/2) = arctan(1) = pi/4
913797
You can begin by converting sin x^2 to 1-cos2x and since they have the same denominator x^2 just split it in two integrals (1/x^2) - (cos2x/x^2).\nThe first one is easy to solve, do initegration by parts for the second integral.
242020
i'll do it by 2 ways \nthe first one if u know what integration by parts is:\n(int=integral)\nint[udv]=uv-int[vdu]\nlet u=lnx =>du=dx/x\n& let dv=xdx => v=(x^2)/2\nso we'll have\nint[xlnx]=((x^2)/2) lnx -1/2int[xdx]\n=((x^2)/2)lnx -((x^2)/4)+ Constante\nthe second method is\ntry derivating (x^2)lnx u'll get [(x^2)lnx]'=2xlnx +x\nthen integrate int{[(x^2)lnx]dx}=(x^2)lnx\n&int[x]=(x^2)/2\nso u'll get (x^2)lnx=2int[xlnxdx]+(x^2)/2\nso int[xlnxdx]=((x^2)/2)lnx-(x^2)/4+ C
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
691116
x/[a^(-x^2/(2a))] dx\nRewrite bringing the denominator to the top.\nThis will change the exponent to positive.\nx(a^(x^2/(2a)) dx\nLet u = x^2/(2a), then du = (1/a)x dx or \na du = x dx\nNow the integral is\nint(a(a^u) du)\n= int(a^(u+1) du)\n= a^(u+1) + C\n= a^(x^2/(2a) + 1) + C\n(The exponent of a is x^2/(2a) + 1)
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
820456
I believe Integration by Parts would be your best bet here, since (a) it's straightforward to differentiate Arcsin(x), and (b) it's not hard to integrate x^(-2).\n\nLet u = Arcsin(x) and let dv = x^(-2) dx\nThus du = 1/sqrt(1-x^2) dx and v = -1/x\n\nSet up the IBP formula:\n\n Int [Arcsin(x)/x^2 dx] = uv - Int [v du]\n  =  Arcsin(x)/x - Int [-1/(x sqrt(1-x^2)) dx]\n  = Arcsin(x)/x + Int [ x^(-1) (1-x^2)^(-1/2) dx]\n... (by substitution*)\n  =  Arcsin(x)/x + ln(x) - ln (sqrt (1-x^2) + 1)\n\nBe SURE to visit the Integrator at Wolfram.com (see link below)!\n\n* See also worked-out example #6 at:\nhttp://mathforum.org/mathtools/cell/c,15.14.6,ALL,ALL/
706596
You'd have to use the chain rule for the square root, then the quotient rule for what's inside.\n\nf ' (x) = (1/2) [(2x³ - 7) / (4x + 1)]^(-1/2) times the derivative of the radicand.\nFor that, the quotient rule gives:\n[6x²(4x + 1) - 4(2x³ - 7)] / (4x + 1)²\n\nThere's most likely some simplifying to be done, but that should get you on the right path.\n\nGood luck!
753779
When you have written it out twice you can then multiply out the brackets\ni.e. x^2(x^2-1/x^2)-(1/x^2)(x^2-1/x^2)\nthis will give you x^4-2+1/x^4\nRewrite using a negative indice to replace the fraction\nx^4-2+x^(-4) and then integrate as per norm using the rule\nthat the integral of x^n = x^(n+1)/(n+1).\nHope your day improves.