whats the integral of 1/((a-x)^2) with respect to x?
127237
let a-x=u\nso, -dx=du\n\nthen ∫1/(a-x)^2dx= \n-∫1/[(u)^2]du=\n-∫u^(-2)du=\n\n-[u^(-2+1)]/(-2+1)=\n\n-u^(-1)/-1=\nu^(-1)=\n\n1/u + C=\n\n1/(a-x) + C
499529
substitute a^1/4 instead of x in the eq.\nu will get\n1-(a^(1/4)^2)=-8\n9=a^1/2\na= 81
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
266972
I assume you are asking for the inverse of f(x)\n\nThen f-1(x) = (x+1/3)/2
650516
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
512937
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
627723
Let u = cx+d, then du = c dx and so (1/c) du = dx\nSolve for x\nu = cx + d\nu - d = cx\nx = (u-d)/c\nThe new integral is\n[a*((u-d)/c) + b]/u (1/c) du\n= (1/c)[(a/c)u - ad/c +b]/u du\n= (1/c)[(a/c) - ((ad/c) + b)(1/u)] du\n= (1/c)[(a/c)u - ((ad/c) + b)ln(u)] + K, where K is a constant\nReplace u with cx+d for your final answer.
516921
Okay, assuming you mean:\n∫e^(2x)/√(1-e^(2x)) dx\nLet u=1-e^(2x), du=-2e^(2x) dx, dx = -1/2 e^(-2x)\n-1/2 ∫1/√u du\n-1/2 ∫u^(-1/2) du\n-u^(1/2) + C\n-√(1-e^(2x)) +C
421068
Distribute the -1/2 through the parentheses and you get a \n-1/4 integral e^2x sin x\n\nAdd this to both sides and you will get the 5/4 integral of e^2x sinx
353839
x/(x+3) - 2(x-1)/(x+3)(x+2)\n\nx(x+2) - 2 (x-1)\n--------------------\n   (x+3) (x+2)\n\n\nx^2 +2\n----------\n(x+3)(x+2)
494268
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
396253
integral(x,-1,2) - 2 * integral(|x|,-1,2)\nintegral(x,-1,2) - 2 * ( integral(-x,-1,0) + integral(x,0,2))\n\n3/2 - 2( 1/2 + 2)\n3/2 - 3\n-3/2
394434
x² / √(1-x) \n= x² √(1-x)/(1 - x)\n= (-1 + 1 - x²)√(1-x)/(1 - x)\n= (1 - x²)√(1-x)/(1 - x) - √(1-x)/(1 - x)\n= (1 + x)√(1-x) - √(1-x)/(1 - x)\n= x√(1-x) + √(1-x) - 1/√(1-x)\n\nSo ∫x² / √(1-x) . dx\n= ∫[x√(1-x) + √(1-x) - 1/√(1-x)] . dx\n\nNow ∫x√(1-x)dx \n= x * -1 * 2/3 (√(1-x)³ - ∫[-1 * 2/3 (√(1-x)³ . 1)dx\n= -2x/3 * √(1-x)³ + 2/3∫(√(1-x)³ dx\n= -2x/3 * √(1-x)³ + 2/3*2/5 * -1 * √((1-x)^5)\n= -2x/3 * √(1-x)³ - 4/15 * √((1-x)^5)\n\n∫√(1-x) dx = \n\n∫1/√(1-x) dx = -2 *√(1-x)\n\nSo ∫x² / √(1-x) . dx\n= ∫[x√(1-x) + √(1-x) - 1/√(1-x)] . dx\n=  -2x/3 * √(1-x)³ - 4/15 * √((1-x)^5) - 2/3 * √(1-x)³ + 2 *√(1-x) + c\n= - 4/15 * √((1-x)^5) -2/3 (x + 1)√(1-x)³ + 2 *√(1-x)  + c
441375
int1/x^2+int1/x^5\nx^-1/-1 +x^-4/-4+C\n-1/x-1/4x^4+C
501864
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!
498430
1 it can be written as sqrt(2) x^-1/2\nSo the integral is 2 sqrt(2) x^1/2\nFrom 1 to 8  f(x) = 2 sqrt(2) [sqrt(8)- sqrt(1)]\n                     f(x) = 2 sqrt(2) [ 2 sqrt(2)-1]\n                     f(x) = 8 - 2 sqrt(2)\n2 it can be written by sinx /cos²x so  -u'/u²\nso the integral is 1/cos x\nFrom -pi/3 to pi/3  f(x) = 1/cos(pi/3) - 1/cos(-pi/3) = 0\n3 The integral is 1/5 arctg (5x) because arctg(x) ' = 1/(1+x²)\n4 The integral is 1/4 tg(4x) because tg(x) ' = 1/cos²(x)
183907
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.