What is the d/dx of e^u?
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
897617
This is an equation with separated variables. Rewrite as \n\ndu/u^3 = (t^3-t)dt \n\nand integrate (find antideritive) of each side separately.
886303
U=sqrt(x) = x^(1/2)\n\ndU= (1/2)*x^(-1/2)dx=(1/2)/(x^(1/2))*dx  (negative exponents on the top of a division symbol can be made positive by moving it to the bottom of the division symbol.)\n\nsolving for dx = (x^(1/2)/(1/2))dU\n\nso sin(sqrt(x)) = sin(U)dx\n\nSo right now our equation looks like\n\n(sin(U)/sqrt(x))*dx\n\nplug in dx and we get\n\nsin(U)*(x^(1/2)/(1/2))dU/sqrt(x)\n\nx^(1/2) = sqrt(x) by the way... so cancel them\n\nsin(U)/(1/2)dU simplified is 2sin(U)dU\n\nnow just integrate that \n\n-2cos(U) + C\n\nplug in U\n\n-2(cos(sqrt(x))) + C
644742
I'm assuming that you are asking for the Laplace transform of the constant "e", or ~2.72. Since the Laplace Transform is only defined for functions that are valid for x > 0, we can write e*u(x), where u(x) is the unit step function.\nOne of the properties of the Laplace Transform is L*a*f(x) is a*F(s), where f(x) is any function of x, a is any constant, L is the Laplace Transform Operator and F(s) is the Laplace Transform of f(x).\nThe Laplace transform of u(t) is 1/s, therefore the transform of the constant "e" is e/s
1183660
Use integration by parts.\nu = 4x, du = 4dx\ndv = e^-xdx, v = -e^-x\n\nint(udv) = uv - int(vdu), so\nint(4xe^-x)dx = -4xe^-x - int(-e^-x(4dx) + c\n= -4xe^-x + 4int(e^-x)dx + c\n= -4xe^-x - 4e^-x + c\n= -4e^-x(x+1) + c
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
749046
well the term you've quoted is a constant...\n\nso (e^2/1+e^2)x would be the answer to that question.\n\nbut i'm guessing you typed it wrong, or are confused.
758976
The total differential is [∂F/∂x]dx +[ ∂F/∂y]dy\n\n∂F/∂x = 4x^3*y^-1\n∂F/∂y = x^4*(-y^-2) = -x^4y^-2
28492
Let u=2x\ndu=2dx\n\nSubstitute in dx=du/2 and u=2x:\nintegral_of( csc2x dx ) = (1/2)integral_of( csc(u) du )\n\nMultiply numerator and denominator by: csc(u)+cot(u)\n==>(1/2)integral_of( [csc^2(u)+csc(u)cot(u)] du/\n[csc(u)+cot(u)] )\n==>(1/2)integral_of( (-1)[-csc^2(u)-csc(u)cot(u)] du/\n[csc(u)+cot(u)] )\n\nLet: v=csc(u)+cot(u)\ndv = [-csc(u)cot(u)-csc^2(u)] du\n\nSubstitute in the above:\n==>(1/2)integral_of(  (-1)dv/v  ) \n==>(-1/2)ln|v| + C\n==>(-1/2)ln|csc(u)+cot(u)| + C\n==>(-1/2)ln|csc(2x)+cot(2x)| + C
980558
∫xe^x dx \n= ∫x d(e^x)/dx dx \n= xe^x - ∫e^x (dx/dx) dx\n= xe^x - ∫e^x dx\n= xe^x - e^x + c\n= (x - 1)e^x + c
664110
consider y=uv where u = x and v =(1-x²)^(1/2).square root is power of half, you should know this.\n\n(dy/dx)=u(dv/dx)+v(du/dx) basic principle that is well known\n(dv/dx)=(0.5)(2x)(1-x²)^(-0.5)\n=x/[(1-x²)^(0.5)]\n(du/dx)=1\nso\n(dy/dx)=x²/[(1-x²)^(0.5)]+[(1-x²)^(0.5)]\n=(x²+1-x²)/[(1-x²)^(0.5)]\n=1/[(1-x²)^(0.5)]\n\nso dy=dx/[(1-x²)^(0.5)]\n\np/s:its very hard to type it in the internet explorer, I could show you a clearer explaination using microsoft word because it has math editor. I hope you will understand, message me if you can't figure out what I'm showing up there
714362
Let u be 3e^x + 1, then du/dx = 3e^x .  So your integral amounts to \n\nintegral( (1/3 du/dx) u^(1/3)) = 1/3 integral(u^(1/3)du)\n\n= (1/3)*1/(1+(1/3))*u^(1 + 1/3) + C\n=(1/3)*(1/(4/3))*u^(4/3) + C\n= (1/3) * (3/4) * (3e^x+1)^(4/3) + C\n= (1/4)* (3e^x + 1)^(4/3) + C\n\nCheck by  differentiating:\n\nd/dx (1/4 (3e^x + 1)^(4/3) + C) = \n(4/3)* (1/4) * (3e^x + 1)^(1/3) * (d/dx (3e^x + 1)) + d/dx(C) =\n(4/12) * (3e^x + 1)^ (1/3) * 3e^x dx + 0 + 0 =\n(1/3) * (3e^x + 1)^(1/3) * 3e^x dx =\ne^x (3e^x + 1) dx\n\nwhich is what you started with.
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
736107
If u = e^x, then 1/u = 1/e^x = e^(-x)\n\n(u - 1/u) / (u + 1/u) = -t\n\nMultiply both sides by (u + 1/u)\n\nu - 1/u = -t (u + 1/u) = -ut - t/u\n\nMultiply both sides by u\n\nu^2 - 1 = -u^2t - t\nu^2 + u^2 t = u^2(1 + t) = 1- t\nu^2 = (1 - t)/(1 + t)\nu = +/-sqrt[(1-t)/(1+t)]\n\nRe-substitute u=e^x:\n\ne^x = +/-sqrt[(1-t)/(1+t)]\n\nTake ln of both sides:\n\nx = ln[sqrt[(1-t)/(1+t)]\n\nSorry I screwed that up the first time.  I need sleep.  Thanks for getting it right, Sidarth.
676902
Incomplete question. Are you looking for the anti-derivative of x, a, or b???\n\nput dx, da, or db at the end of the equation so we'll know.
871386
∫udv = uv - ∫vdu\n∫x^2dx/√(1-x)\nlet u = x^2, dv = dx/√(1-x)\ndu = 2xdx, v = -2√(1-x)\n∫x^2dx/√(1-x) = -2x^2√(1-x) + ∫(4x√(1-x))dx\nu = 4x, dv = √(1-x))dx\ndu = 4dx, v = -(2/3)(1-x)^(3/2)\n∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) + ∫(8/3)(1-x)^(3/2)dx\n∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) - (16/15)(1-x)^(5/2)\n∫x^2dx/√(1-x) = √(1-x)(2x^2 - (8x/3)(1 - x) - (16/15)(1-x)^2)\n∫x^2dx/√(1-x) = √(1-x)(2x^2 - 8x/3 - 8x^2/3 - (16/15)(1 - 2x + x^2)\n∫x^2dx/√(1-x) = (1/15)√(1-x)(30x^2 - 40x - 40x^2 - 16 + 32x + 16x^2)\n∫x^2dx/√(1-x) = -2√(1 - x)(x(3x + 4) + 8)/15
703729
You want to start by integrating the equation. Fortunately, this is not difficult.\n\ndy/dx = (x-2)e^(-2y)\n\nInt(e^(2y)dy) = Int((x-2)dx)\n\n (1/2)e^(2y) = (1/2)x^2 - 2x + C, where C is some constant\n\ne^(2y) = x^2 - 4x + C  Note that e^(-2y) = 1/[x^2 - 4x + C]\n\n2y = ln (x^2 - 4x + C)\n\ny = (1/2)ln (x^2 - 4x + C)\n\nWe are almost done, now we just need to plug in the given condition, and solve for C.\n\nln (2) = (1/2)ln(4 - 8 + C)\n2ln(2) = ln(C-4)\n4 = C - 4\nC = 8\n\nSo the particular solution is y = (1/2)ln (x^2 - 4x + 8)\n\nIf you want to check this, we need to differentiate it\n\ndy/dx = (1/2)(2x-4)/(x^2 - 4x + 8)\n\nRemember from above that  e^(-2y) = 1/[x^2 - 4x + 8]\n\nSo dy/dx = (x-2)e^(-2y) as required