what is the integral of  √(3+x^2)      and √(3+x^3) ?
414992
√(3+x^2) = INt √(√3)^2 +X^2)\n\nBy formula INt √(a^2 +x^2) = x{√(a^2 +x^2)}/2 +a^2 log {x +√(a^2 +x^2)}/2 +c\n\nAns:- x√(3+x^2)/2 + 3/2 *log {x + √(3+x^2)} + c
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
822637
When I have problems with an indefinite integral I use some of my symbolic software to get straight to answer or a "Table of Integrals".  And then I focus on how to get to that particular answer, whether it is through substitution, changing it to a double integral in polar coordinates, etc.\n\nA "free" online integrator program that is offered by Wolfram (maker of Mathematica) is located at\n\nhttp://integrals.wolfram.com/\n\nType your particular problem as\n \nArcTan[1/x] \n\nand it will give you an answer that is solved symbolically for an "indefinite" integral which is\n\nx(arctan(1/x)) + (1/2)(log((x^2) + 1))\n\nHowever to be honest with you it was a few years back when I saw this particular integral, and I honestly don't remember the process.  I think it was a definite integral though.
819535
sqrt(x) - sqrt(3x + 3) = 1\nsqrt(x) = sqrt(3x + 3) + 1\n\nsquare both sides\n\nx = (sqrt(3x + 3) + 1)^2\nx = (sqrt(3x + 3) + 1)(sqrt(3x + 3) + 1)\nx = (3x + 3) + sqrt(3x + 3) + sqrt(3x + 3) + 1\nx = 3x + 3 + 2sqrt(3x + 3) + 1\nx = 3x + 4 + 2sqrt(3x + 3)\n\n-2x - 4 = 2sqrt(3x + 3)\n-x - 2 = sqrt(3x + 3)\n\nsquare both sides\n\n(-x - 2)^2 = 3x + 3\n(-x - 2)(-x - 2) = 3x + 3\nx^2 + 2x + 2x + 4 = 3x + 3\nx^2 + 4x + 4 = 3x + 3\nx^2 + x + 1 = 0\n\nx = (-b ± sqrt(b^2 - 4ac))/(2a)\n\nx = (-1 ± sqrt(1^2 - 4(1)(1)))/(2(1))\nx = (-1 ± sqrt(1 - 4))/2\nx = (-1 ± sqrt(-3))/2\n\nx = (1/2)(-1 ± isqrt(3))\n\nThere are no real solutions.
737583
A bit tricky to integrate, but is like this (i could mistake!):\n\na)\n9^(1/5)\nS pi ( (3 sqrt(x))^2 - x^3^2) dx\n0\n\nb)\n9^(1/5)\nS pi ( (3 sqrt(x)-6)^2 - (x^3-6)^2) dx\n0\n\nc)\n9^(3/5)\nS pi ( (y^2/9)^2 - (y^(1/3))^2) dy\n0\n\nd)\n9^(3/5)\nS pi ( (y^2/9-4)^2 - (y^(1/3)-4)^2) dy\n0\n\nRegards
38734
(2/3)x^3 + (5/2)x^2\nyou just take the antiderivitive of each term.  so you backtrack. instead of taking away one from the power, you add one and then divide the whole term by the new power. so the first term would be to the third power. now if you took the derivitive you would multiply by that power of three, so we need to cancel that out by dividing by three. so it's (2/3)x^3  because when you take the derivitive and that three comes down, it will be (3)(2)/(3) and the 3's will cancel out giving you 2x^2 which is what you started with. so the antiderivitive is (2/3)x^3.\n\nnow we do the second term. so we add a power, so it would be squared. now we have to cancel that power out so we'll divide the whole thing by two(or rather, multiply by a half) so the antiderivitive of 5x would be (5/2)x^2. this can be checked by finding the derivitive. multiply the (5/2) times 2 and you are left with 5 and take a power away to get 5x.  integrals are easier than you think, its just the opposite of differentiation. the general formula is:  X^n integrated is (X^n+1)/n+1 so if it's x^3 it would become x^4/4 or (1/4)x^4. again, you can check this by differentiating it. the exponent of 4 is multiplied and cancels out the fraction, and the exponent turns to a 3 and you're left with what you started with...\nhope you understand integrals now :)
872911
Let Z = 3 - 3i\n|Z| = √(3² + 3²)\n= 3√2\nThus Z = 3√2(1/√2 - 1/√2i)\nNow swwk an angle such that cosθ = 1/√2 and sinθ = -1/√2\nVlearly this is a 4th quadrabt angle\nTherefore θ = 2π - π/4\n= 7π/4\nThus Z = 3√2(cos7π/4 + i sin7π/4)\n[= 3√2e^(i7π/4)]
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.
879050
You need to make the trig substitution, x = 2 tan(u). Then dx = 2 sec^2(u) du.\nThe integral becomes\n(1/(4tan^2(u)*sqrt(4tan^2(u)+4)))*2sec^2(u)\n\nUsing the Pythagorean identity, tan^2(u) + 1 = sec^2(u), the square root disappears and stuff cancels. If you rewrite everything in terms of sin and cos, more stuff cancels. You should get\ncos(u)/(4sin^2(u))\n\nA better way to write this is\n(1/4)(cot(u))(csc(u))\nsince this is very close to the derivative of csc(u), which is\n-cot(u)csc(u)\n\nThus the integration is -(1/4)csc(u).\nIf you draw a right triangle with an angle with measure u, the leg opposite u has length x, the leg adjacent to u has length 2, since we said above that x=2 tan(u), or equivalently, tan(u)=x/2.\nThe hypothenuse of this triangle is sqrt(x^2+4), by the Pythagorean theorem. Then csc(u) is sqrt(x^2+4)/x.\n\nThe final answer is -sqrt(x^2+4)/4x.\n\nI tried this on the Mathematica Integrator and the answer it gave matches mine. You have to input 1/(x^2*Sqrt[x^2+4]). Note that Sqrt is capitalized, to tell Mathematica that you mean Sqrt, the function, not sqrt, the variable; also note the use of square brackets (instead of rounded parentheses), which indicate the thing you're square-rooting.
1143501
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
97125
use the distance formula\nsqrt(3^2+(-sqr3)^2)=?
1092276
Let u = 3 - x^2, then du = -2x dx\nso (-1/2) du = x dx\nNow the integral is\n(-1/2)(1/u^2) du\n= (-1/2)(u^(-2)) du\n= (-1/2)u^(-1)/(-1) + C\n= (1/2)(u^(-1)) + C\n= 1/(2u) + C\n= 1/(2(3-x^2)) + C\nThere is no need to solve for x = sqrt(-u+3).  The x in the integral cancels with the x from du.
498269
2-root-3 is the same as 2 ^ 1/3 (^ is exponent sign)\n\n2 ^ 1/3 squared is (2 ^ 1/3) ^ 2 ... you multiply the exponents, so it equals\n\n2 ^ 2/3
281867
Well I assume you are trying to work out x and I also assume that it's not cos [to the power of] (x^3) but Cos [multiptied by] (x^3)\n\nYou need to integrate to remove the d/dx:\n\nCos integrated is Sin and you then have to divide by the differental of what's in the brackets\n\n=> sin(x^3) / 3x^2
864682
∫x³/√(x² + 9)dx Let x = 3 tanθ therefore dx = 3 sec²θ . dθ\n=∫27 tan³θ/√(9 tan²θ + 9) . 3 sec²θdθ\n= 27 tan³θ/√(tan²θ + 1) . sec²θ . dθ (tan²θ + 1 = sec²θ)\n= 27∫tan³θ/secθ . sec²θ . dθ \n= 27∫tan³θsecθ . dθ \n\n= 27∫tanθsecθ (sec²θ + 1) . dθ \n= 27∫(sec²θ. tanθsecθ + tanθsecθ) . dθ (NOTE dsecθ/dθ = tanθsecθ)\n= 27( ⅓sec³θ + secθ) + c\nIf x = 3 tanθ then tanθ = x/3 so cosθ = 3/√(x² + 9)\nhence secθ = √(x² + 9)/3\n\nThus required integral = ⅓(x² + 9)^(3/2) + 9(x² + 9)^½ + c
910252
1 it can be written as sqrt(2) x^-1/2\nSo the integral is 2 sqrt(2) x^1/2\nFrom 1 to 8  f(x) = 2 sqrt(2) [sqrt(8)- sqrt(1)]\n                     f(x) = 2 sqrt(2) [ 2 sqrt(2)-1]\n                     f(x) = 8 - 2 sqrt(2)\n2 it can be written by sinx /cos²x so  -u'/u²\nso the integral is 1/cos x\nFrom -pi/3 to pi/3  f(x) = 1/cos(pi/3) - 1/cos(-pi/3) = 0\n3 The integral is 1/5 arctg (5x) because arctg(x) ' = 1/(1+x²)\n4 The integral is 1/4 tg(4x) because tg(x) ' = 1/cos²(x)
907483
1/x^2 = x^(-2)\nThe antiderivative is x^(-2+1)/(-2+1) + C\n= x^(-1)/-1 + C\n= -1/x + C\nIt is similar for 1/x^3\nYou do partial integrals the same way you do regular integrals.  You need to know which one you are integrating with respect to and then pretend the other variable(s) are constants.