what is the sub integration?
541630
I believe it refers to the people who go to live and work to the USA from another country (in this particularly context), but who doesn't really integrates him or her self to the culture, costumes and traditions of the country where they have migrate to, and remains with their original countries tradicions, culture and costumes - in some cases even conserve the languaje.
403929
Integration by parts is done by breaking down an integral so that\n\nS u dv = uv - S v du (where S is the integration symbol)\n\nFor example, if we were to integrate xcos x dx,\n\nlet u = x\ndu = dx\n\nthen dv = cos x (all the other stuff we didn't use for u)\nv = sin x (just integrate dv)\n\nand the formula will look like\n\nS xcos x dx = xsin x - S sin x dx\n= xsin x + cos x + C\n\nIt's kinda an art form or experience in picking which part of the integral to make u and dv - there's really no rule that says u = cos x.  Experience (or trial and error) will show you that that's probably not the better choice though .
963003
int(y+2/2y^2+3y+1 dy)\n\n1/2*y^2 - 1/y + 3/2*y^2 + y + c\n\n2*y^2 - 1/y + y + c
473554
You have to use integration by parts.\nLet u = ln(x) and dv = dx, then\ndu = (1/x)dx and v = x\nint(udv) = uv - int(vdu)\n= xln(x) - int(x(1/x)dx)\n= xln(x) - int(dx)\n= xln(x) - x + C, where C is a constant
507439
a sub has a crossover that only lets lows go to the sub(bass only)   ,while your mids and highs go to the other speaker.for a much clearer crisp sound.  .....and that sue nami guy dont now what hes talkin about!
105440
You can try and find constants A and B such that\n\n1/(x(x-a)) = A/x + B/(x-a)\n\nIf you get this, then the integration is simple.\n\nHow do you get it?\n\nStart from A/x + B/(x-a) and calculate the common denominator:\n[A(x-a) + Bx]/[x(x-a)] = [(A+B)x - aA]/(x(x-a))\n\nNow to get the expression equal to 1/(x(x-a)) it must be:\n\nA+B = 0  ---> A= -B\naA = 1  ---> A = 1/a\n\nso \nA = 1/a (constant)\nB= - 1/a\n\nsubstitute and you get the integral as the sum of two simpler ones which get two logarithms by integration.\n\nThen you apply the property of the sum of logarithms to obtain a single one.\n\nI hope it's clear.
274471
i made a little comment for you on your integral answer:)
1122081
It's called the integration by parts formula.\nInt u dv=uv- int v du\n\nBoth u and v need to be absolutely continuous for this to work. A continous derivative for both is plenty.
887107
B.the electron subshell
760474
simple\n\nintegration = addition \ndifferentiation = division
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))
586082
integrate by parts.\n\nintegral(lnx dx)\n= (lnx)(x) - integral (x d(lnx))\n= xlnx - integral (x (1/x) dx)\n= x lnx - integral (1 dx)\n= x lnx - x + c
197427
Sub Main()\n        Console.WriteLine(2 ^ 5)\n        Console.ReadLine()\nEnd Sub
763570
in calculus, integration is just an extension of sum. for discrete values, we use the summation. for continuous values, we use integration.\n\nindefinite integration is just integrating indefinitely, ie., without any bounds. this usually results in a polynomial or trigonometric equation etc. \n\ndefinite integration is done between two specific points, be it on the x axis or y axis. this results in a constant value, rather than a polynomial or other equations.\nthe applications of definite integration is to find the area under curves, etc..
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/
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
74089
http://integrals.wolfram.com/