how to intergrate this integral-----S radical(1+lnx) / xlnx?
410987
You were on the right track with the u substitution:\n\nlet u = sqrt(1 + ln x)\nthen u^2 = 1 + ln x\nu^2 - 1 = ln x\n2u du = dx/x\n\nSubstitute in to the original integral:\nS 2u^2/(u^2 - 1) du = 2S u^2/(u^2 - 1) du\n\nI found in an old book that I have that\nS x^2/(x^2 - a^2) dx = x + (a/2)ln[(x - a)/(x + a)]\n(I know it works, but I can't seem to explain how my book came up with that one.  To prove it to yourself, take the derivative of the right side and it works out to the expression inside the integral on the left.)\n\nSo, \n= 2[u + (1/2)ln[(u – 1)/(u + 1)]\n\nAnd substituting for u,\n= 2sqrt(1 + ln x) + ln[(-1 + sqrt(1 + ln x))/(1+ sqrt(1 + ln x))]
748051
The primitive of your integral is:\n\ns*(e^(st))*(1/w)*atan(s/w)\n\nYou have to do a trigonometric substitution to get there. Do not always trust the weird answers you get with complex calculators.
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
631241
the integration by part is very long and complicated\ni used it many times\nlet u =logsinx          dv=dx\n   du= cosx/sinx          =x\nxlogsinx-int{xcosx/sinx}\nint{xcosx/sinx}  by sbutitution\nlet u=sinx   x=sin^-1(x)\ndu=cosx dx dx=du/cosx\nyou will get int{sin^-1(u)/u} remmeber u=sinx\nby part\nlet k=1/u          dv=sin^-1(u)\n     -1/u^2        int{sin^-1(u)}by part\n\nint{sin^-1(u)}\nlet m=sin^-1(u)                dv=du\n     =1/sqt(1-u^2)                u du\n=usinx^-1(u)-int{u/spt(1-u^2)}\nint{u/spt(1-u^2)} int this by subitution\n\nthen u will get\nsinxsin^-1(x)+(1/2)(1-sin^2x)\n\nso the final ans\nxlogsinx-sinxsin^-1(x)\n+(1/2)(1-sin^2x)+c\n\nplz try to answer this by yourself ,i hope i gave you the idea,thankx
849119
Let u = -x^2.\n\ndu = -2xdx.\n\n-1/2 du = xdx\n\nSo Integral [x*e^(-x^2)]dx = Integral[-1/2 e^u]du\n\nWith the substitution done, I think you should be able to evaluate the integral and back-substitute.
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
147855
Try http://integrals.wolfram.com/
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
789574
from the ⌠udv = uv - ⌠vdu, let u=(lnx)^2 and dv=x^4dx\nhence du=2dx/x and v=x^5/5\nsubstituing values we have:\nx^4 (lnx)^2 dx = (lnx)^2(x^5/5) - ⌠(x^5/5)(2dx/x)\n                      = (lnx)^2(x^5/5) - ⌠(2/5)x^4dx\n                      = (lnx)^2(x^5/5) - x^5/10\nthen using the limits; this yields to 1.796442507
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.
157812
[x^2/4][2(log(x) - 1) log(x) + 1]
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
14407
This is an integration-by-parts problem. The answer is:\nx*ln(x)+C\n\nThis is how you calculate it:\n\nIntegral of ln(x)*dx\n\nu= ln(x)\ndu=1/x*dx\ndv=dx\nv=x\n\nSo by parts: uv-integral of vdu\n= ln(x)*x-integral of x*1/x*dx\n=x*ln(x)-integral of 1*dx\n=x*ln(x)-x+c
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.
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))
682208
int[(x^3 - x - 3sqrt2)/(x^2 - 2)]\nFirst use long division\n(x^3 - x - 3sqrt2)/(x^2 - 2)\n= x + (x - 3sqrt2)/(x^2 - 2)\nThen use partial fraction decomposition to rewrite\n(x - 3sqrt2)/(x^2 - 2)\nx^2 - 2 = (x-sqrt2)(x+sqrt2)\n(x - 3sqrt2)/(x^2 - 2) = A/(x-sqrt2) + B/(x+sqrt2)\nx - 3sqrt2 = A(x+sqrt2) + B(x-sqrt2)\nPlugging in x = sqrt2 gives A = -1 and plugging in \nx = -sqrt2 gives B = 2\nThe integral is now\nint[x - 1/(x-sqrt2) + 2/(x+sqrt2)]\n= (1/2)x^2 - ln(x-sqrt2) + 2ln(x+sqrt2) + C