Integration question...?
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
1106006
Let u = cx+d, then du = c dx and so (1/c) du = dx\nSolve for x\nu = cx + d\nu - d = cx\nx = (u-d)/c\nThe new integral is\n[a*((u-d)/c) + b]/u (1/c) du\n= (1/c)[(a/c)u - ad/c +b]/u du\n= (1/c)[(a/c) - ((ad/c) + b)(1/u)] du\n= (1/c)[(a/c)u - ((ad/c) + b)ln(u)] + K, where K is a constant\nReplace u with cx+d for your final answer.
245944
The integral of a function over a definite interval.
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 -.......
424492
Redo the integral.  \nFor #1 I got pi*(3pi-4)/2\nint(2pi*sin(x)((3/4)pi-x))\n= 2pi*int((3/4)pi*sin(x) - x*sin(x)\n= 2pi*[-(3/4)pi*cos(x) - sin(x) + x*cos(x)]\nFor pi/2:\n2pi[-(3/4)pi*0 - 1 + (pi/2)*0]\n= -2pi\nFor 0:\n2pi*[-(3/4)pi*1 - 0 + 0]\n= 2pi(-3/4)pi\n= (-3/2)pi^2\nNow\n-2pi - (-3/2)pi^2\n= -2pi + (3/2)pi^2\nYou can simplify this to the above answer by factoring out pi and then getting a common denominator of 2.  (It probably isn't necessary though)\n-2pi + (3/2)pi^2\n= pi*(3pi-4)/2\n\nFor 2, I got pi*(pi-2).\nThe integration will be similar to the one above, so I won't type it out again.  You can message me if you have a problem with the second integration.
452294
Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
955693
i don't understand the way you\nhave written down the function\n\nsay that you want to integrate\n2x/(2x-1)\nput u=2x-1,>>>x=(u+1)/2,\ndx=du/2,\nint{2x/(2x-1)}dx\n=int{(2*(u+1)/2)/u}*du/2\n=1/2*int{(u+1)/u}du\n=1/2*int{1+1/u}du\n=1/2[u+lnu]+C\nbut u= 2x-1\n=1/2[2x+ln(2x-1)-1]+C\nwhere C is a constant\n\ni hope that this helps
1143501
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
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.
880226
it is not what you specified\n\nIt cannot be solved without identities\n\nsin ^5 t = sin^4 t sin t\n= (1-cos^2 t)^2 sin t\n\nnow let cost = x\nso diffentiate to get - sint dt = dx\n\nso we need to integrate\n\n-(1-x^2)^2 dx\n\n= -dx + 2 x^2 dx - x^4 dx\n\nintegrating we get\n\n-x + 2x^3/3 - x^5/5 + C\n= - cos t + 2 cos^3 t/3 - cos^5 t /5 + C\n\nwhere C is constant of integration
847056
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps
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
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
117158
the integration >>> increase the power by one and divide by the new power (don't forget the the constant)\n\nx^n -------> x^(n+1)/(n+1)\n\ny = 1/(x^2)\ny = x^-2\nthe integration = x^(-1)/(-1) + c\n= -x^-1 + c\n= -1/x + constant
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.
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
700700
try putting x^2 = i*t\nwhere i = complex iota\n t is varible \n\nthere integral will be in form : i/2*( cos(t)+i*sin(t) ) dt\n\nso answer will be : 1/2 * (i* sin (t) + i*cos(t) ) = 1/2 * e^ (i*t) = 1/2 * e^x^2\n\nThat's the answer....... :D\n\n\n[Assuming u know that e^(i*t) = cos(t)+i*sin(t) ]\n\ncan be solved easily ......... try it