antiderivative of sqrt(1+4x^2)?
598163
The answer is\n\n(1/2)(x sqrt(4x^2+1)+ (1/2)ln(2x+ sqrt(4x^2+1)))\n\n\nheres the formula: int(sqrt(a^2+x^2))=\n(x/2)(sqrt(x^2+a^2))+\n(a^2/2)(ln(x+sqrt(x^2+a^2)))\n\nmight look complicated but take it down in a piece of paper and it will be pretty easy..
372419
I fear you have a missing element here... do you want the differencial of 4P^2..etc because if so you need to know what 4P^2..etc equals, or is 4P^2..etc the differencial of another expression, because if so your question is a bit like saying what is the answer to x=2\n\nIf you want to differencial of 4P^2..etc then I suggest you solve the original equation for t which will make the algebra easier and then use the chain rule where dP/dt = [1/(dt/dP)]
641733
[-b +/-  sqrt(b^2-4ac)]/(2a)
1165906
1/sqrt(1-v^2/c^2), I think
781653
Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
920193
e^(xlnx )=x^x\nderivative is  1+ln x
866864
Just differentiate it. You need f'(x) < 0.\n\nBy the quotient rule, f'(x) = (2x(x^2 + 4) - (x^2)(2x))/(x^2+4)^2.\nNow, since the bottom is a square, thus positive, you just need the numerator to be < 0:\n2x(x^2 + 4) - x^2 * 2x < 0\n2x^3 + 8x - 2x^3 < 0\n8x < 0\nx < 0\nThus it is decreasing on (-∞,0).
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)
889651
It's just arctan (x^2).\n\nBy the fundamental theorem of calculus, if F(t) is the antiderivative of f(t) = arctan(t^2), then the integral of f(t) from 0.5 to x will give F(x) - F(0.5). The derivative of this function with respect to x then gives you f(x).
821024
∫cos(x^2) dx = √(π/2)FresnelC(√(2/π)x)
169299
Completing the square is a technique of elementary algebra wherein an expression x2 + bx is replaced by one of the form x + c)2 + d. \n\nA simple example is this.\n\nx2 + 4x = (x + 2)2 − c = (x2 + 4x + 4) − 4 \n\nNow, consider the problem of finding this antiderivative:\n \nThe denominator is 9x2 − 90x + 241 = 9(x2 − 10x) + 241. \nAdding (10/2)2 = 25 to x2 - 10x gives a perfect square x2 - 10x + 25 = (x - 5)2. So we get\n\n9(x2 − 10x) + 241 = 9(x2 − 10x + 25) + 241 − 9(25) = 9(x − 5)2 + 16. \n\nCompleting the square reduces any problem involving a quadratic polynomial to one involving a square quadratic polynomial and a constant.
852487
all are possible\n\nanttderivative is x^2/2+3x +c\n1st one ok\n2nd ok ok\n3rd one x^2/2+3x + 9 ok\n4th one x^2/2+3x -8\nall differe only by a constant\n\nso all are possible
1156233
The dist is\nsqrt.[(4-3)^2+(1-1)^2+(4-3)^2]=sqrt.2
289224
(x+ [1+i*sqrt(3)]/2) * (x+[1-i*sqrt(3)]/2) = 0
586893
If its differentiate x * sqrt 1 + x^2 , the answer is\n  1 + 2x ( since sqrt 1 is a constant and d x^2/dx = 2x)\n\n \nIf your question is sqrt1 + x^2 answer is 0 + 2x = 2x
764329
x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.