Find the INT of 1 / (x^2+4)^2?
482644
Let us denote the integral sign by I.\nWe will use substitution method.\nLet x = 2*tan(t)\nThen dx = 2*sec^2(t)dt\nThus, I[1/(x^2+4)^2]dx\n= I[1/{4*tan^2(t)+4}^2]*2*sec^2(t)dt\n= (1/8)*I[1/{tan^2(t)+1}^2]*sec^2(t)dt\n= (1/8)*I[1/{sec^2(t)}^2]*sec^2(t)dt  [since tan^2(t) + 1 =sec^2(t)]\n= (1/8)*I[1/{sec^2(t)}]dt\n= (1/8)*I[cos^2(t)]dt\n= (1/16)*I[2*cos^2(t)]dt\n= (1/16)*I[1+cos(2t)]dt [since 2*cos^2(t) = 1 + cos(2t)]\n= (1/16)*I[dt] + (1/16)*I[cos(2t)]dt\n= (1/16)*t + (1/16)*[{sin(2t)}/2] + C where C = arbitrary constant of integration\n= t/16 + {sin(2t)}/32 + C\n\nNow sin(2t) \n= {2*tan(t)}/{1+tan^2(t)}\n= 4x/(4+x^2) [since x = 2*tan(t)]\n\nAlso t = tan-1(x/2) [tan-1(x/2) means tan inverse (x/2)]\n\nHence the given integral equals to\n{tan-1(x/2)}/16 + x/{8(4+x^2)} + C
857154
sure.\n\nx ^ 4 - a ^ 4 = (x ^ 2 - a ^ 2) * (x ^ 2 + a ^ 2)\n\nthen for x no equals to a or - a :\n\n(x^2-a^2) / (x^4-a^4) = 1 / (x ^ 2 + a ^2) -------> 1 / a ^ 4
171290
I think you mean "integral" of x^(n-1)* exp(-(log)^2)/2 and I THINK that the e and the log should cancel so you would have X^(n-1) * -X^2/2 = x^n-1+2 /2 so the integral of that would be... X^(n+1)/(2n+4)
1092533
The previous person almost got there.\nThe indefinite integral is (1+x)[log(1+x) - 1] + C\n\nSubstitute x=1 to get 2[log(2) - 1] + C\nSubstitute x=0 to get  -1 + C.\n\nThe definite integral is \n2[log(2) - 1]  - [-1]\n= 2log(2) -1\n=log(4) - 1\n=log(4) - log(e)\n=log(4/e)
629659
x^4 + 200 =102x^2\n   x^4 -102x^2 + 200 = 0        Now split the middle term\n   x^4 -100x^2 - 2x^2 + 200 = 0 Factor the first two and last two terms\n   x^2(x^2-100) -2(x^2-100) = 0\n   (x^2-2)(x^2-100) = 0\nEither (x^2-2) = 0  or (x^2-100) = 0\n        x^2 = 2     or  x^2 = 100\n        x= sqrt(2), -sqrt(2)    or x= +10, -10
801052
It seems you are asking for f'(2)?  in which case, we have\nf'(x) = 1/(1+x^4)so that f'(2) = 1/9\n\nSteve
283258
I=int dx/[(x^2)*(x^2+1)^.5]\n\nLet x=tan u\n   dx=secu^2 du\n\nSubtitute\n\nI=int (secu^2 du)/[(tanu^2)*(tanu^2+1)^.5]\n\nBut tanu^2+1=secu^2\n\nThen \nI=int (secu^2 du)/[(tanu^2)*(secu^2)^.5]\n--> \nI=int (secu^2 du)/[(tanu^2)*(secu)]\n-->\nI=int (secu du)/[(tanu^2)]\n\nBut\nsecu=1/cosu ..... and tanu=sinu/cosu\n\nThen\nI=int (1/cosu)/(sinu/cosu)^2\nI=int (1/cosu)*(cosu/sinu)^2\nI=int cosu/(sinu)^2\nI=int (cosu)*(sinu)^-2\nI= -(sinu)^-1\nI= -cscu + c .......... c is a constant\n\n\nBut tanu=x --> sinu=x/[(x^2+1)^0.5]\n\nthen\n-cscu= -1/sinu= -1/[x/(x^2+1)^0.5]= -[(x^2+1)^0.5]/x\n\nthen\nI = -[(x^2+1)^0.5] / x + c
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
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)
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......
297440
Are you looking for indefinite  integral   of x^(sin(1/x)) ?\n\nMathematica has a web site you can plug-in the integrant and get the result (if it exist in closed form)\n\nBy hand: try to break this power expression using log and exp functions. If  a=ln(b), then b=exp(a) and therefore b=exp(ln(b)).\n\nIn this way your integrant could be written as \n\nexp(ln(integrand))\nand since it is of power form, use the properties of ln and exp functions to break it down and thus reduce it to simpler integrals.
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
984538
(2/6) / (4/9)\n1/3 * 9/4\n3/4
806158
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
784758
x^2+x+1=0\nx^2+2(1/2)x+1=0\nX^2+2(1/2)x+(1/2)^2+1-(1/2)^2=0\n(x+1/2)^2+3/4=0\n(x+1/2)^2=-3/4\nx=-1/2+/-√(3/4) i=-1/2+/-√3 * i/2
682208
int[(x^3 - x - 3sqrt2)/(x^2 - 2)]\nFirst use long division\n(x^3 - x - 3sqrt2)/(x^2 - 2)\n= x + (x - 3sqrt2)/(x^2 - 2)\nThen use partial fraction decomposition to rewrite\n(x - 3sqrt2)/(x^2 - 2)\nx^2 - 2 = (x-sqrt2)(x+sqrt2)\n(x - 3sqrt2)/(x^2 - 2) = A/(x-sqrt2) + B/(x+sqrt2)\nx - 3sqrt2 = A(x+sqrt2) + B(x-sqrt2)\nPlugging in x = sqrt2 gives A = -1 and plugging in \nx = -sqrt2 gives B = 2\nThe integral is now\nint[x - 1/(x-sqrt2) + 2/(x+sqrt2)]\n= (1/2)x^2 - ln(x-sqrt2) + 2ln(x+sqrt2) + C
678157
2x^2(1-2x)=0\n\nx=0\nx= (1/2)