Integration question?
337080
when you integrate 1/x you get In |x|\n\nso the answer is -1/2 * In |x| + constant term\n\n\nYou can use substitution, but there's not much point, because you'll get 1/u then (which is the same thing). 1/x is a well known integrand. You'd be better just memorising that the integral of 1/x = In(x)
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!
574666
int((x-2)*(x-5)^(1/2)*dx)\n\nUse u sub, let:\nu=x-5\ndu=dx\nx-2=u+3\n\nResub:\n\nint((u+3)*sqrt(u)*du)\n\nDistribute:\n\nint((u^3/2 + 3sqrt(u)*du))\n\nIntegrate:\n\n0.4*u^(5/2) + 2*u^(3/2) + c\n\nResub u for x:\n\n0.4*(x-5)^(5/2) + 2*(x-5)^(3/2) + C
94569
Your integral needs either limits (from 1 to X, for example) or a constant added to it. With limits, the apparent contradiction vanishes. With an arbitrary constant, it can be accounted for.
68076
god man.  what kind of horrible painful calculus class are you taking?\n\nfirst integrate over dx.  the integral of x^y dx is just a power integral, so you have y*x^(y-1).  the integral of y^x dx is just an exponential, so you have y^x/ln y.\n\nnow integrate y*x^(y-1) + y^x/ln y over dy.  and there, my friend, is where i stop because frankly i can't be arsed to do something so brain-curdling.  actually the first term doesn't look so bad, i think if you integrate it by parts it should come out in one step.  don't really want to think about doing the 2nd one.\n\nyou might consider trying some kind of fancy thing like transforming to polar coordinates or something.  might make it easier.\n\nbtw you should know about this site:\nhttp://integrals.wolfram.com/
808989
Let t=sinx\n=> dt = cosx dx\n\nx=pi/2 => t=1\nx=0 => t=0\n\nI=integrate(e^t dt) from 0 to 1\n\n=e^t from 0 to 1\n\n=> I= e^1 - e^0 = e-1\n\nAnswer is I=e-1
744903
1)\n\ntan^2(x) = sec^2(x) - 1\n\nSo the integral of sec^2(x) + 4 = tan(x) + 4x\n\n2)\n\ncsc(x) = 1/sin(x)\n\nthe integral is then of 1+sin(x)/cos^2(x).\nlet u = cos(x) then du = -sinx dx.\nso the integral is x+sec(x).\n\n3)\n\nlet u = cos(2x+1) then du = -(1/2)sin(2x+1) dx.\n\nso the integral is of -(1/2)/u^2 = 2/u = (1/2)/cos(2x+1) = sec(2x+1)/2\n\nSteve
414817
int = integral\nint[(sinx)^5dx]\nRewrite (sinx)^5 as sinx (sinx)^4\n(sinx)^4 = ((sinx)^2)^2\n(sinx)^2 = 1 - (cosx)^2\nThus, \n(sinx)^4 = ((sinx)^2)^2\n= (1-(cosx)^2)^2\nNow the integral is\nint[sinx(1-(cosx)^2)dx]\nUse substitution.\nLet u = cosx, then du = -sinxdx\nNow the integral is\nint[-(1-u^2)^2du] \nExpand (1-u^2)^2 and multiply through by the -1\n-(1-u^2)^2 = -(1 - 2u^2 + u^4) = -1 + 2u^2 - u^4 \nThe integral now equals\nint[(-1 + 2u^2 - u^4 )du]\n= -u +(2/3)u^3 - (1/5)u^5 + C, where C is a constant\nReplace u with cosx to get the final answer.\n-(cosx) +(2/3)(cosx)^3 - (1/5)(cosx)^5 + C
147855
Try http://integrals.wolfram.com/
704235
I would expand (sin 2x+1)^2, then replace (sin 2x)^2 by\n(1-cos 4x)/2, then integrate term by term.  If you can't do that, I expect another answerer will do the whole thing, or else you can email h_chalker@yahoo.com.au
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
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
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
775488
Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.
772619
integral(x,-1,2) - 2 * integral(|x|,-1,2)\nintegral(x,-1,2) - 2 * ( integral(-x,-1,0) + integral(x,0,2))\n\n3/2 - 2( 1/2 + 2)\n3/2 - 3\n-3/2
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.