a. ln |x^3-6x^2|+C\nb. 1/3 ln |x^3-6x^2|+C\nc. 3 ln |x^3-6x^2|+C\nd. cant be integrated using formulas
199089
S (x^2-4x) / (x^3-6x^2) dx\n\nlet u = x^3-6x^2\nthen du = 3x^2-12x\n\nS (x^2-4x) / (x^3-6x^2) dx = (1/3) du/u\n= (1/3) ln u + C\n= (1/3) ln (x^3-6x^2) + C\n\nAnswer B.
314949
1/2 x^2 -3x\n- 2/3 x^2 +3x -5\n________________\n\n-1/6x^2 (3x's cancel out) -5\n______________________________\n\n-1/6x^2 -5
315248
a) (.033)^3\nb) (.008)^3\nc) (.066)(.027)(.033)
489303
ln(x) + 2ln(4) = ln(7) - ln(x)\nln(x) + ln(4^2) = ln(7) - ln(x)\nln(x*4^2) = ln(7/x)\nln(16x) = ln(7/x)\n16x = 7/x\n16x^2 = 7\nx^2 = 7/16\nx = sqrt(7/16)\nx = sqrt(7)/4
374387
Assuming that x^4 is not in the denominator:\n\n(x-2/x^2) simplifies to  (x^-1) - (2x^-2)\n\nTherefore, the problem becomes:\n\nint[ x^-1  -  2x^-2  + x^4 ]\n\nNow you can use the power rule to integrate each term:\n\nln(x) + 2x^-1 + (x^5)/5  +  C
279042
a.  ax + bx\nb.  21x^3 + 24x^2y\nc.  24x^2 + 31x + 10\nd.  24x^2 - 8x - 6\ne.  40x^3 + 66x^2 + 38x + 6  (I assumed that you meant 7x in the first half of the equation.  If not, this is not correct.)\nf.  x^2 -1\ng.  x^3 - 8
498452
a.x=[-8+/-rt(64-24)]/2\n\nb.x=[-5+/-rt(25-56)]/4\n\nc.x=[-7+/-rt(49-64)]/4\n\nd.x=[-5+/-rt(25-72)]/4
384935
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
591559
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
376210
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
540349
(6x^2)/(4x^2 y-12xy) * (x^2 + x - 12)/ (3x^2 + 12x)\n\nFirst, I would attempt to simplify if possible:\n2x(3x)/x(2xy-6y) * (x^2 + x - 12)/ (3x^2 + 12x )\n(3x) / (2xy - 6y) * (x^2 + x - 12)/ (3x^2 + 12x)\n\nRewrite to see if we can cross cancel:\n\n(3x)           (x^2 + x - 12)\n---------  *   -----------------\n(2xy-6y)    (3x)(x + 4x)\n\n1               (x^2 + x - 12)\n---------- *  -----------------\n(2xy-6y)     (x+4x)\n\nUnfoil the (x^2 + x -12 to see if we can further simplify\n\n(x^2 + x - 12) = (x -3)(x+4) so you will have:\n\n1             (x-3)(x+4)\n-------  *  --------------\n2y(x-3)      (x+4)\n\nBy simplification:\n\n1           (x-3)\n------   * ----------\n2y(x-3)     1\n\nBy Cross Cancel:\n1/2y
656547
=(x+3)(x-1) - (x-2)(x+4)  /  (x+4)(x-1)\n=(x^2 +2x - 3) - (x^2 + 2x - 8) / (x^2 + 3x - 4)\n=(5) / (x^2 + 3x - 4)
490952
We know that F(x) = ∫ln³ x/x dx + C. So we need to find that integral, solve for the constant of integration, and then plug 9 into the resulting formula. First, the integral:\n\n∫ln³ x/x dx\nLet u=ln³ x, v=ln x, du=3 ln² x/x dx, dv = 1/x dx. Integrating by parts:\n∫ln³/x dx = ln^4 x - ∫3 ln³ x/x dx + C\nAdding 3∫ln³ x/x dx to both sides:\n4∫ln³/x dx = ln^4 x + C\nDividing by 4:\n∫ln³/x dx = ln^4 x + C\n\nNow, solving for C:\n\nF(1) = ln^4 1 + C\nF(1) = 0^4 + C\nF(1) = C\nBut F(1) = 0, so C=0.\n\nNow, solving for F(9):\n\nF(9) = ln^4 9 = (ln 9)^4 ≈ 23.30761
515724
you should start with the power series representation of ln(1+x), because this is one of the standard ones that is either given or you have memorized\n\nln(1+x) = x - (1/2)x^2 + (1/3)x^3 - (1/4)x^4 + . . . \n\nso ln(4-x) is what you get when you sub in 3-x in for x in the above equation\n\nln(4-x) = (3-x) - (1/2)(3-x)^2 + (1/3)(3-x)^3 - (1/4)(3-x)^4 + . . . \n\nand now, xln(4-x) is when you multiply the entire series above by x\n\nxln(4-x) = x(3-x) - x(1/2)(3-x)^2 + x(1/3)(3-x)^3 - x(1/4)(3-x)^4 + . . .
494955
(A) 3+2(x-1)\n\n3+2x+2\n2x+5\n\n(B) 5-2(x+3)\n\n5-2x-6\n-2x-1\n\n(C) 1+2(x-1)\n\n1+2x-2\n2x-1\n\n(D) 6-3(w-2)\n\n6-3w+6\n-3w+12
47386
(x+3)(x^2-x+4)/(x+3) common factor\n\nmultiply by x+3\n\nx^2-x+4
409868
ln (x-1) + ln (x-5) = ln(3x+7)\nln( (x-1)(x-5) ) = ln(3x+7)\nso\n(x-1)(x-5)  = (3x+7)\nx^2 - 6x + 5 = 3x + 7\nx^2 -9x -2 =0\nx= (9+- sqrt( 81 -4(-2)) )/2\nx = ( 9 +- sqrt(89) )/2\n`