evaluate: (x^2+5)/(x^3+15-2x^2+1)^2 dx on [0,1] find the antideriviative with the fund theo of cal?
110508
(x^2+5) / (x^3+15-2x^2+1)^2\non [0,1]
809242
-ln(15)/ln(5)... i just used the solve function on my calculator\n\nsolve(5^x=1/15,x)
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
506639
The first derivitive of f(x) = 5/x^3 - 3/x^1/3 is...\n\nf'(x) = -15/x^4 + 1/x^(4/3)\n\nHope this helps. I used the product rule. Remember, you can write the term 5/x^3 and 5*x^-3. I find it simpler to rearrange the equation and use the product rule to get at the first derivitive.
841476
When you retry the problem remember to move only yout z^-2 up don't move the 4 that is a constant in the denominator. \n\nyou should get this . (12x^5z^3/30x^7y^7) * (15x^2z^10/4y^3)\n\nremember to simplify as best you can prior to multiplying.\n\ntry from here and let me know what you get.
545883
(a^2  - 5) * (a^2  - 2)
842823
You start by finding an antiderivative of each one (not difficult, just break up the fraction and use the power rule on each part). Then you evaluate it at the limits of integration, and subtract. For example:\n\n[1, 8]∫(6x²+2)/√x dx\n[1, 8]∫6x^(3/2) + 2x^(-1/2) dx\n12/5 x^(5/2) + 4x^(1/2) from 1 to 8:\n12/5 (8)^(5/2) + 4(8)^(1/2) - (12/5 (1)^(5/2) + 4(1)^(1/2))\n1536√2/5 + 8√2 - (12/5 + 4)\n1576√2/5 + 32/5\n\nYou can do the second one on your own.
317595
If f is an antiderivative of (tanx)^2/(x^2+1) such that f(1)=0.5, then f(0)=__?\n\nnumerical integral of (tan x)^2/(x^2 + 1) from 0 to 1 = 0.345\nf(x) = numerical integral of (tan x)^2/(x^2 + 1) from 0 to x + 0.155\nso f(0) = 0.155
771554
(x+5)(x-5)\n(x)^2-(5)^2\n=x^2-25
919338
Well, I see you've used the correct order of operations this time. That's good - it shows that you're improving.\n\n#1: Apply the product and chain rules to find f'(x), then evaluate at 0:\n\nf'(x) = d(√(x³+5x+121))/dx (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * d(x³+5x+121)/dx * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * (3x²+5) * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nNow evaluating :\n\nf'(0) = 1/(2√121)*5*11 + √121 = 5/2 + 11 = 27/2\n\n#2: First, evaluate the function at y=1 to get the value of x:\n\nxy^2 -2y + 4y^3 = 6\nx - 1 + 4 = 6\nx = 3\n\nNow, implicitly differentiate both sides:\n\nd(xy^2 -2y + 4y^3)/dx = d(6)/dx\ny² + 2xy dy/dx - 2 dy/dx + 12y² dy/dx = 0\n\nSolve for dy/dx:\n\ny² + (2xy - 2 + 12y²) dy/dx = 0\n(2xy - 2 + 12y²) dy/dx = -y²\ndy/dx = -y²/(2xy - 2 + 12y²)\n\nNow evaluate at (3, 1):\n\ndy/dx = -1/(6 - 2 + 12) = -1/16\n\n#3: First, find the derivative:\n\nf'(x) = (x²+x-2 - (2x+1)x)/(x²+x-2)² (using the quotient rule)\n\nSimplifying:\n\nf'(x) = (x²+x-2 - 2x² - x)/(x²+x-2)²\nf'(x) = (-x²-2)/(x²+x-2)²\n\nNow we must find where f'(x) < 0\n\n(-x²-2)/(x²+x-2)² < 0\n-x²-2 < 0 (we know (x²+x-2)² is nonnegative, and if the expression on the left is defined, it is also nonzero. Thus it is positive and we can multiply by it without changing the inequality)\n-x² < 2\n\nNow, since x is a real number, this expression will always be true. Thus, if f'(x) is even defined, x is decreasing. Now f'(x) is defined provied that (x²+x-2)²≠0. So:\n\n(x²+x-2)²≠0\n(x-1)²(x+2)²≠0\nx≠1 and x≠-2\n\nWhich makes sense, since those are vertical asymptotes of the original function. Thus, the function is decreasing on:\n\n(-∞, -2) ∪ (-2, 1) ∪ (1, ∞)
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)
38332
(x-2)/(x^2-2x-15)+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(-2-x)(x+3)/(x-5)(x+3)\n\n[(x-2) + (x+3)(-2-x)] / [(x+3)(x-5)]\n\nANSWER :\n(x^2+4x+8)/((x+3)(5-x))
802168
The function is of the form d(u/v) which is given by\n(v du - u dv) / v^2.\nhere u=(1+x)^2, hence du=2(1+x)\nhere v=1-x, hence dv=-1.\nsubstitute these values in the formula and simplify to get final result.\n\nKSR
349210
dy/dx = -(15cos(3/x)/2x^2) * (sin(3/x))^3/2
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
76994
The absolute value function is just:\n\n{ x < 0    | -x\n{ x >= 0 | x\n\nso the antiderivative is:\n\n{ x < 0    | -x^2 / 2 + c\n{ x >= 0 | x^2 / 2 + c
750140
For 7), take the derivative of the numerator and denominator separately.\n\nlim as x -> 5 of (2x)/1\n\nthe limit is 10.\n\nCheck: for x = 5.01, the function evaluates to 10.01.\nfor x = 4.99, the function evaluates to 9.99.\n\nEasier: factor the numerator into (x + 5)(x - 5) and cancel the x - 5. That leaves you with x + 5 FOR ALL VALUES OF X, not just the limit.\n\nGopal's solution to 7) is wrong because Gopal cannot add 5 + 5.\n\nI can't read 8), and it contains a backward slash, which is not a mathematical symbol. 8) looks wrong. Why have you expressed a constant as 5(.5)^4 instead of multiplying it out? Anyway, taking the derivative of the numerator and denominator separately gives you\n\n20(.5 + h)^3\n\nAs h goes to zero, the limit is 20(.5)^3\n\nGopal's solution for 8) is wrong because Gopal took the derivative wrong.\n\nAre you sure you wrote 9) correctly?\n\nFor 10) to be differentiable at x = 2, the slopes of the two functions must be equal at x = 2, whether you approach from the left or from the right.\n\nfor x <= 2, f'(x) = 3ax^3 + 5\nAt x = 2, f'(2) = 24a + 5\n\nFor the second function, what does "bx2" mean?\n\nFor the functions to be continuous, they must evaluate to the same value at x = 2. The first function evaluates to 16a + 10 at x = 2. The second function is anyone's guess.