evaluate the integral of [ (1+x)/(1-x) ]^1/2 dx?
463304
(sinx)^(-1) - sqrt(1-x^2) + C\nStart by multiplying (1+x)/(1-x) by (1+x) on the top and bottom\n[(1+x)(1+x)]/[(1-x)(1+x)]\nMultiply the denominator out but not the numerator.\n[(1+x)(1+x)]/[(1-x)(1+x)] = [(1+x)(1+x)]/(1-x^2)\nNow\nsqrt[(1+x)/(1-x)] \n= sqrt[[(1+x)(1+x)]/(1-x^2)]\n= (1+x)/sqrt(1-x^2)\nNow use trig substitution again letting x = sin(u)\nthen dx = cos(u) du\n(1+x)/sqrt(1-x^2) dx \n= [(1+sin(u))/sqrt(1-(sinu)^2)]cos(u) du\n= [(1+sin(u))/sqrt((cosu)^2)]cos(u) du\n= [(1+sin(u))/cos(u)]cos(u) du \nthe cosines cancel\n= [1 + sin(u)]du\n= u - cos(u) + C\nNow x= sin(u) so u = (sin x)^(-1)\nand using a triangle you can see that cos(u) = sqrt(1-x^2)
202670
There is no closed form of this integral. Expressions of this form are power towers, and you can use series methods to evaluate this integral. The value of this integral can be found in the source link. :)\n(To a poster above: All integrals are real, it's just that some may not be expressible in closed form.)
793695
' the integral of 1/e^u du = integral of e^{-u}du\n=-e^{-u} + C
955693
i don't understand the way you\nhave written down the function\n\nsay that you want to integrate\n2x/(2x-1)\nput u=2x-1,>>>x=(u+1)/2,\ndx=du/2,\nint{2x/(2x-1)}dx\n=int{(2*(u+1)/2)/u}*du/2\n=1/2*int{(u+1)/u}du\n=1/2*int{1+1/u}du\n=1/2[u+lnu]+C\nbut u= 2x-1\n=1/2[2x+ln(2x-1)-1]+C\nwhere C is a constant\n\ni hope that this helps
949467
You divide each term in the upper part by x^2 to get a simple polynomial ((10 x^9) / x^2) + ( 5 / x^2) = 10 x^7 + 5 x^-2\nThe integral is obtain by the rule Integrate(x^n) = (x^(n+1)) / (n + 1)
796927
try the link below...... hope it can put an end to ur integral problems
147855
Try http://integrals.wolfram.com/
734906
the integral of [cos 2x dx] is [1/2 sin 2x + c]
738333
(1+sin^2x)dx/\n(1+sin^2x)(1-sin^2x)\n=dx/(1-sin^2x)\ndx/cos^2x\nsec^2xdx\ntanx+C
757602
[(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)
753779
When you have written it out twice you can then multiply out the brackets\ni.e. x^2(x^2-1/x^2)-(1/x^2)(x^2-1/x^2)\nthis will give you x^4-2+1/x^4\nRewrite using a negative indice to replace the fraction\nx^4-2+x^(-4) and then integrate as per norm using the rule\nthat the integral of x^n = x^(n+1)/(n+1).\nHope your day improves.
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
822637
When I have problems with an indefinite integral I use some of my symbolic software to get straight to answer or a "Table of Integrals".  And then I focus on how to get to that particular answer, whether it is through substitution, changing it to a double integral in polar coordinates, etc.\n\nA "free" online integrator program that is offered by Wolfram (maker of Mathematica) is located at\n\nhttp://integrals.wolfram.com/\n\nType your particular problem as\n \nArcTan[1/x] \n\nand it will give you an answer that is solved symbolically for an "indefinite" integral which is\n\nx(arctan(1/x)) + (1/2)(log((x^2) + 1))\n\nHowever to be honest with you it was a few years back when I saw this particular integral, and I honestly don't remember the process.  I think it was a definite integral though.
491455
(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
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)
963003
int(y+2/2y^2+3y+1 dy)\n\n1/2*y^2 - 1/y + 3/2*y^2 + y + c\n\n2*y^2 - 1/y + y + c
888834
Integrate from -π/2 to π/2 and double it