find the integral x^2-4x/x^3-6x^2 using substitution method?
443446
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.
985787
this is\n\nx^4+ 4\n= (x^2)^2 + 2^2\nadd and subtract 4x^2\n= (x^2)+2(x^2)*2 + 2^2 - 4x^2\n= (x^2+2)^2 - (2x)^2\n= (x^2+2x+2)(x^2-2x+2)
739098
Move the constant to the other side of the equation:\n2x^2 -6x = 11\nDivide out the x^2 coefficient:\nx^2 - 3x = 11/2\nTake half of the coefficient of x, square it, and add it to both sides of the equation:\n(-3/2)^2 = 9/4\nx^2 -3x + 9/4 = 31/4\nThe left-hand side of the equation can now be expressed as a square where the constant added to x is the number you just squared and added to both sides:\n(x - 3/2)^2 = 31/4\nTake the square root of both sides, remembering that the right side must be expressed in plus-minus form:\nx - 3/2 = +/- sqrt 31/4\nMove the constant on the left side of the equation to the right in order to isolate the variable:\nx = 3/2 +/- sqrt 31/4\nSimplify.  Since sqrt (31/4) is the same thing as (sqrt 31)/(sqrt 4), you have:\nx = 3/2 +/- (sqrt 31)/2\nx = (3 +/- sqrt 31)/2
801525
This is a tricky one. You have to use a couple of steps to get this one. First you use a u-substituion to "simplify" it and then use a trig substitution.\n\nIf you let u=cos(x) then du=-sin(x) then the integral becomes\n&#92;int -1/(1+u^2) du.\n\nNow let u=tan(s) and du=sec^2(s) and the integral is\n&#92;int -sec^2(s)/(1+tan^2(s)) ds = &#92;int -sec^2(s)/sec^2(s) ds = &#92;int -1 ds.\n\nThe integral is -s + C where C is any constant. We now have to undo the steps to get it back in terms of x. First, we undo the trig substituion. Since u=tan(s) then the integral is\n\n-arctan(u)+C\n\nSince u=cos(x) we get\n\n-arctan(cos(x))+C
421543
use substitution:\n\nlet u=-x^2, then du= -2x dx, so 4x dx= -2du \n\nso now we have the integral of -2e^u wrt to u\nwhich is:\n\n-2e^u\n\nnow putting back in that u=-x^2\n\nwe have the answer:\n\n-2e^(-x^2)
414104
Using an integral,\n\nWords = S 3t(3t^2 + 1)^(-1/2) dt\n\nlet u = 3t^2 + 1\nthen du = 6t dt\n\n= (1/2) S u^(-1/2) du\n= (1/2)(2u^(1/2))\n= u^(1/2)\n= (3t^2 + 1)^(1/2)\n\nEvaluating from 1 to 4,\n= (3(4)^2 + 1)^(1/2) - (3(1)^2 + 1)^(1/2)\n= 49^(1/2) - 4^(1/2)\n= 7 - 2\n= 5\n\nAnswer A.
820806
-2x^2+6x-3=0\n=->2(x^2-3x+3/2)=0\n=>2(x^2-2x.3/2+3^2/2^2)-3/4=0\n=2(x+3/2)^2-(sqrt3/2)^2=0\n=>-2(x+3/2+sqrt3/2)(x+3/2-sqrt3/2)=0\nTherefore,x=-3/2-sqrt3/2 or -3/2+sqrt3/2
966462
(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
581691
if by this you mean\n\n((x^2)/2) - ((7x)/4) = (-3/2)\n\nMultiply everything by 4\n\n2x^2 - 7x = -6\n2x^2 - 7x + 6 = 0\n(x - 2)(2x - 3) = 0\nx = 2 or (3/2)\n\nANS : x = 2 or (3/2)
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)
781368
By substitution\ny = x^2\ndy = 2x dx\n\nint x2^(x^2) dx = (1/2) int 2^y dy = (1/(2ln2))2^y \n= (1/(2ln2))2^(x^2)
346785
(x^2-5x+4)/(x^2-9) * \n(x^2+7x+12)/(x^2+3x-28)=\n\n[(x^2-5x+4)*(x^2+7x+12)]/\n[(x^2-9)*(x^2+3x-28)]=\n\n[(x^2-4x-x+4)*(x^2+4x+3x+12)]/\n[(x-3)(x+3)*(x^2+7x-4x-28)]=\n\n[(x(x-4)-(x-4))*(x(x+4+3(x+4)]/\n[(x-3)(x+3)*(x(x+7)-4(x+7)]=\n\n[(x-4)(x-1)*(x+4)(x+3)]/\n[(x-3)(x+3)*(x+7)*(x-4)]=\n\n[(x-1)(x+4)]/[(x-3)(x+7)]
709461
2x^2-6x+3\nuse this formula\n-b+or-root b^2-4ac/2a\nIn this case \na=2   b=-6    c=3\njust substitute this value in above formula\nYour answer must be\n3+root3/2     3-root3/2
849348
There are two methods, Method 1 you could simplify (1-X^4)^7 then multiply through by X^3, but... you dont want to do that, especially on an exam. \n\nMethod 2 is substitution.\n\nLets make up an equation here that we can substitute into our original equation...\n\nSay U=(1-X^4)  so we have (X^3)U^7dx, understand?\n\nNow lets take the derivative of U....  so dU= -4x^3dx, and solving for dx = dU/(-4x^3)\n\nLets plug that into our equation for dx.\n\nwe now have (X^3)*U^7dU/(-4x^3)\n\nCanceling the X^3 were left with (-1/4)*U^7dU\n\nNow you can integrate that right!\n\n(1/8)*(-1/4)*U^8 \n\nfurther simplified is\n\n(-1/32)U^8\n\nNow just plug back in U=(1-X^4) and we get\n\n(-1/32)*(1-X^4)^8\n\nThat should be the right answer!
904086
I would use substitution method on this one...\n\nFirst, break it up into the sum of 2 integrals...\n\nInt(cost/(1+sin^2t))dt + Int(-2sint/(1+cos^2t))dt\n\nFor the first integral let u = sint\n\ndu = cost\n\nThen you have an integral that you can evaluate ...\n\nInt(1/(1 + u^2)du = arctan(u)\n\nSubstitute cos(t) back in for u and you get arctan(cos(t)), evaluate that from 0 to pi/2.\n\nFollow the same reasoning for the 2nd one!
747340
-2x-x > -6 + 4\n-3x > -2 \nx < 2/3
191567
Please use parentheses to explain the problem more effectively. Assuming the problem as,\n\n(x-1)/2x + x/(x+3)\n\nThe solution can be,\n\n(3x^2+2x-3)/(2x^2+6)