why isn't integration of x^2 dx from 1 to 10  equal to 1^+2^+3^.....10^?
970210
LHS is a summation at all points continuosly where as RHS is a sumamtion at descrete points/
438371
same thing as (8x^2+2x-2)^(3/2), right?\nso add one to the power\nmultiply by the inverse of that power\nand divide by the derivative of what's in the parenthesis on the left\ngot it?
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/
769880
The exponent is not divided by (1 + 1) in 2*x^(1+1)/(1+1), the whole expression is. ie 2x²/2 = x². Its not to be confused with 2x^(2/2) ie 2x^1 = 2x\n\nWhen there are limits it makes no difference\n\neg ∫(from x = a to x = b) 2x dx \n= [x² + c](from x = a to x = b)\n= (b² + c) - (a² + c)\n=  b² - a² (irrespective of the value of c)
781368
By substitution\ny = x^2\ndy = 2x dx\n\nint x2^(x^2) dx = (1/2) int 2^y dy = (1/(2ln2))2^y \n= (1/(2ln2))2^(x^2)
677393
it is probably easiest to expand into a polynomial like bboy does. or use u substitution:\nlet u =1-3x\ndu=-3 dx\nso integral of -2/3 u^2 = -2/9u^3 = -2/9(1-3x)^3
1122457
1st question:\ny = x+10\ndy = dx\n\nint((y-6)e^(y-10)/y^7)dy = (e^-10)int(e^y/y^6)dy - 6(e^-10)int(e^y/y^7)dy\n\nTo the side (for integration by parts for the first integral on the right side):\nu = 1/y^6, du = -6/y^7dy\ndv = e^ydy, v = e^y\n\n= e^(-10)((e^y/y^6) - int(e^y)(-6/y^7)dy) - 6(e^-10)int(e^y)/(y^7)dy +c\n= e^-10(e^y/y^6) + 6e^-10int(e^y/y^7)dy - 6e^-10int(e^y/y^7)dy + c\n= e^(y-10)/y^6 + c\n= e^x/(x+10) + c\n\nSecond question:\nDifferential on both sides\n\nyx^(y-1)dx + x^y/ln(y)dy + xy^(x-1)dy + y^x/ln(x)dx = 0\ndx(yx^(y-1) + y^x/ln(x)) + dy(x^y/ln(y) + xy^(x-1)dy) = 0\ndy(x^y/ln(y) + xy^(x-1)) = -dx(yx^(y-1) + y^x/ln(x))\ndy/dx = -(yx^(y-1) + y^x/ln(x))/(x^y/ln(y) + xy^(x-1))\n\nYou really ought to ask these as separate questions though.
706536
Integral: ∫ [2/{(e^2x)+4}] dx\n\nFirst consider the expression: 2/{(e^2x) + 4}\n\n= {e^(2x) + 4}/[2*{e^(2x) + 4}] - e^(2x)/[2*{e^(2x) + 4}]\n\n= 1/2 - (1/2)[e^(2x)/{e^(2x) + 4}]\n\n\nHence, ∫[2/{(e^2x) + 4}] dx\n\n=  ∫[(1/2) - (1/2)[e^(2x)/{e^(2x) + 4}]] dx\n\n=  ∫(1/2) dx - ∫(1/2)[e^(2x)/{e^(2x) + 4}] dx\n\n=  (1/2) x - (1/2)∫[e^(2x)/{e^(2x) + 4}] dx\n\nLet {e^(2x) + 4} = z. Then dz = 2e^(2x) dx\n\nHence the given integral becomes:\n(1/2) x - (1/4)∫(dz/z)\n\n= (1/2) x - (1/4)ln|z| + c where c = arbitrary constant of integration\n\nNow, substitute the value of 'z' in terms of 'x', thus the given integral becomes\n= (1/2) x - (1/4)ln|e^(2x) + 4| + c
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
671135
hmmm... I'm getting that the integral of sec(x)^3 is\n  .5sec(x)tan(x)+.5ln|secx+tanx|\n\nsince....\n\nintegral sec (x)^n =  1/(n-1)*sec(x)^(n-2)*tan(x)+\n(n-2)/(n-1)*integral sec(x)^(n-2) \n\nI'm not sure how it could just be ln|sec(x) + tan(x)|\n\n\nit is (sec (x))^3 and not sec(x^3) right? (just thought i'd double check!)
699436
(2^x-2^-x) / 3 (the whole thing is divided by 3)\n\nThat is not an equation to be solved.   So I don't know where you are getting an equals sign (=).??\n\nYou may be able to simplify it, but you cannot say x= something unless you were provided with an EQUATION.\n\nI will do some work on it.\n\nAfter further review, try to input your value for x into the statement\n(2^x-2^-x) / 3 and what you get is another question.   Your value then EQUALS what?   Does this answer your question?    All you will get is a value.
837567
Interesting...there are actually two solutions. Let's use substitution:\n\nLet u = cos(x). du = -sin(x)dx.\n\nSo Integral[sin(x)cos(x)]dx\n\n= - Integral[udu]\n\n= -u^2/2\n\n= -[cos(x)]^2/2\n\nNow, let u = sin(x). du = cos(x)dx.\n\nSo Integral[sin(x)cos(x)dx]\n\n= Integral[udu]\n\n= u^2/2 = [sin(x)]^2/2\n\nJust to check, I did a definite integral on both results, and got the same value for both.\n\nPretty neat...
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
408775
1/(1+e^x)=(1+e^x -e^x)/(1+e^x)=1-  e^x/(1+e^x)=\n1- (1+e^x)'/(1+e^x)\n& u'll integrate
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.
896031
Why not?  If you divide top and bottom of this expression by x^7 you get\n(3 + 4/(x^5))/(9 - 1/(x^4) + 2/(x^7))\n\nEvery term except the 3 and the 9 now has a power of x as denominator, and therefore approaches 0 as x goes to infinity,\ntherefore it approaches 3/9.
591447
Well ... thought I had solved it, but then I got stuck!!!  Integration is not the best thing I am at!!!!\n\nUsing "$" as the integral sine.\n\nUse integration by parts: $udv = uv - $vdu\n\nThen split the orignal question into:\n\ny = $ (2x-3)(1/(√(9-x^2))) dx\n\nu = 2x - 3\ndu = 2\n\ndv = 1 / √(9 - x^2)\ndv = 1 / √(3^2 - x^2)          ==> standard integral\nv = sin^-1(x/3)     \n\nThen use the integration by parts equation above to try and solve it!  \n\ny = (2x-3)(sin^-1(x/3)) - $2 / sin(x/3)\n\nThen you have to solve that integral by parts again (using "z" this time, so not to be confused with the whole equation which is equal to "y")\n\nz = $2 / sin(x/3)\n\nu = 2\ndu = 0\n\ndv = 1 / sin(x/3)\nv = 3[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nz = $2 / sin(x/3) = 6[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nOK, (I hope this is right) so ... then put that back into the above equation that I got after the first round of integrating by parts!\n\nThat is: y = (2x-3)(sin^-1(x/3)) - $2 / sin(x/3)\n\ny = (2x-3)(sin^-1(x/3)) - 6[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nWhich eventually simplifies down to:\n\ny = [(4x - 3)/(sin(x/3)] - 6√(1 - (x/3)^2)]\nor\ny = [(4x - 3)/(sin(x/3)] - √(36 - 4x^2)]\n\nLet me know if that was right, if you ever find out!!!!\n\nI hope it was ... good luck