What is the answer for this indefinite integration?
79944
The above is the derivative.\nIntegration gives x + (1/4)x^4 + c (generic constant).
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......
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
245944
The integral of a function over a definite interval.
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)
263227
This integral does not have an indefinite closed form of elementary functions. It can be evaluated numerically by either integrating the series representation of e^(-x^2) or using integral approximation methods like Simpson's rule.\nIt also has an interesting value if evaluated over its entire domain; by changing coordinates to polar coordinates, one can easily show that the integral over the entire real line is equivalent to Sqrt[Pi].\nThis fact is used in statistics to normalize the Gaussian distribution to have area 1 under the curve. The integral of your expression on the interval [0,t] is then known as the error function. See the source for details.
400264
split the integral up\n\nyou have:\n(1-2x^2)/x = 1/x - 2x^2/x = 1/x - 2x\n\nintegrating this gives:\n\nIn(x) - 2x^2/2 = In(x)-x^2
480262
the integration for any intgral from (a) to (-a) is always equal zero.\nso the intgral for this is zero\nbut if there is no interval the integration is very complicated
786528
Integration is one of the two major operations of calculus (the other being the inverse, differentiation. There are indefinite and definite types of integration. When the integral of a function is expressed symbolically (as another function), this is the indefinite integral (e.g., the ind. int. of y=2x is y=x^2 + a constant). When the integral is numerically evaluated over some range of x, this is the definite integral. Much of time-domain simulation consists of successive evaluation and accumulation of the definite integral of some function, or "time stepping". Typically, each time step starts with evaluation of the 2nd derivative of something, let's say position (whose 2nd deriv. is acceleration), accumulation of the 1st deriv. (velocity) as the previous value + time-step size * 2nd deriv., then accumulation of the 0th deriv. (position) as the previous value + time-step size * 1st deriv. Of course it's a bit more complex than that, and many integration formulas exist to minimize errors.\nAn "integration" function could be something like what I've described for time-stepping, or it could be a much more sophsticated function that accepts a function definition and gives you the indefinite integral in symbolic form. So you need more information about what kind of integration function is desired.
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
759209
This is a non-elementary integral in x. However, from the way you phrased the problem, I would guess that you want this to compute a double integral: first with repspect to x, then with respect to y. The best way of doing *that* would be to switch the order of integration. Then the integral with respect to y will be xe^(y/x). Be sure to change the limits of integration so that you are integrating over the same region in the xy-plane!
894395
∞\n\nConsider that this function is everywhere positive, and that as the lower bound of the integral increases, the function being integrated does not approach zero (in fact, it becomes increasingly large), so the integral of e^(y²) from -∞ to anything must in fact be infinite.
297440
Are you looking for indefinite  integral   of x^(sin(1/x)) ?\n\nMathematica has a web site you can plug-in the integrant and get the result (if it exist in closed form)\n\nBy hand: try to break this power expression using log and exp functions. If  a=ln(b), then b=exp(a) and therefore b=exp(ln(b)).\n\nIn this way your integrant could be written as \n\nexp(ln(integrand))\nand since it is of power form, use the properties of ln and exp functions to break it down and thus reduce it to simpler integrals.
842647
Take the antiderivative (indefinite integral) of little f. This will give you some function of x plus a constant. Then plug zero into the antiderivative, set the whole expression equal to zero (to satisfy F(0)=0) and solve for the constant. Now that you know F(x). Plug in three. This is the answer.
1136463
It's been a while since I had calculus, but I think it's something like this:\n\nYou'd have to take the integral from 0 to k (treat k as a constant). When you have that, take that value (which is equivalent to the area) and set it equal to 3. Then, algebraically, solve for k.
719380
Let y = x^(1/2)\n\nso y^2 = x,\n\ndx = 2y dy\n\nThe problem becomes:\n\nint [1 / (y + y^3)] 2y dy\n\nOr:\n\nint 2 [1 / (1 + y^2)] dy\n\nA recognizable integral. Integrate and substitute back x^1/2 = y
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