How do you integrate 1/(1-Q) with respect to Q?
408576
its a logarithmic integral.\ndifferentiation of log x is 1/x\ntherefore integration of 1/a-x is -log(a-x)+c\nnegative sign because coefficient of x is -1
770190
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
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.
704235
I would expand (sin 2x+1)^2, then replace (sin 2x)^2 by\n(1-cos 4x)/2, then integrate term by term.  If you can't do that, I expect another answerer will do the whole thing, or else you can email h_chalker@yahoo.com.au
376765
Let u = 1 + x^4 and then apply the Chain Rule. This allows your expression to be written as\n\n1/sqrt(u)^(1/2)\n\nwhich is equivalent to\n\n1/u^(1/4)\n\nwhich, in turn, is equivalent to\n\nu^(-1/4).\n\nDifferentiating this expression with respect to x yields the following:\n\n(-1/4)*u^(-1/4-1)*(du/dx).\n\nNote: du/dx = 4x^3.\n\nTherefore, the desired derivative is \n\n(-1/4)*(1+x^4)^(-5/4)*(4x^3).\n\nIf we simplify this a bit we are given\n\n-x^3/(1+x^4)^(5/4).
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
361025
Here, you need the exercise. Show how\n1/(x^2-1) + 1/(x+1) = x/(x^2-1)\nSimple enough fraction. And it's free.
793695
' the integral of 1/e^u du = integral of e^{-u}du\n=-e^{-u} + C
473554
You have to use integration by parts.\nLet u = ln(x) and dv = dx, then\ndu = (1/x)dx and v = x\nint(udv) = uv - int(vdu)\n= xln(x) - int(x(1/x)dx)\n= xln(x) - int(dx)\n= xln(x) - x + C, where C is a constant
521923
Let 'I' denote the integral sign.\nWe will use the method of substitution.\nLet e^2x = z.\nThen taking differentials of both sides 2*e^2x*dx = dz or 2*z*dx = dz or dx = dz/(2*z).\n\nNow 1/((e^-2x)+( e^2x)) can be written as (e^2x)/[1 + (e^2x)^2]\n\nHence given integral = I[(e^2x)/{1 + (e^2x)^2}]dx\n= I[z/(1 + z^2)]{dz/(2*z)} [substituting z = e^2x]\n= 0.5*I[1/(1 + z^2)]dz\n= 0.5*tan-1(z) + C [tan-1(z) denotes tan inverse z]\nwhere C = arbitrary constant of integration\n= 0.5*tan-1(e^2x) + C
847056
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps
246551
integral of sine squared.\nso theres a conversion thing which says \nsin^2 x = (1-cos 2x)/2\nso if you use that then u have to integrate 1/2 - cos 2x/2\nwhich is just 1/2x - sin 2x / 4
607013
Another Method.\n\n-1 = -1\n\n-1/1 = 1/-1\n\nSquare root both sides,\n\nSQR(-1/1) = SQR (1/-1)\n\nExpanding,\n\nSQR(-1)/SQR(1) = SQR(1)/SQR(-1)\n\nSQR(-1)/1 = 1/SQR(-1)\n\nCross Multiply,\n\nSQR(-1)*SQR(-1) = 1\n\n(SQR(-1))^2 = 1\n\n-1 = 1\n\nDivide by 2 on both sides,\n\n-1/2 = 1/2\n\nAdd 3/2 to both sides,\n\n-1/2 + 3/2 = 1/2 + 3/2\n\n1 = 2 (ans)
464039
Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
117158
the integration >>> increase the power by one and divide by the new power (don't forget the the constant)\n\nx^n -------> x^(n+1)/(n+1)\n\ny = 1/(x^2)\ny = x^-2\nthe integration = x^(-1)/(-1) + c\n= -x^-1 + c\n= -1/x + constant
860131
It's undefined at the endpoint of the interval (pi/2).\nFirst go ahead and integrate it like you normally would.\nintegral of sin(x)/sqrt(cosx) dx\nUse substitution.  Let u = cos x, then du = - sinx dx\n= integral of -du/ sqrt(u) \n= integral -u^(-1/2) du\n= -u^(-1/2 +1) / (-1/2 + 1) \n= -u^(1/2) / (1/2)\n= -2u^(1/2)\n= -2 sqrt(u)\nReplace u with cosx\n= -2 sqrt(cos x)\nNow plug in pi/2\n- 2 sqrt(cos pi/2)\n= -sqrt(0)\n= 0\nPlug in pi/4\n-2 sqrt(cos pi/4)\n= -2sqrt( sqrt(2) / 2)\nNow subtract them\n0 - (-2sqrt(sqrt(2) /2))\n= 2sqrt( sqrt(2) / 2)
743452
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