What is integral of log(logx)+1/(logx)^2 ?
147855
Try http://integrals.wolfram.com/
634484
log_2(X) + log_2(X+5) = 1\nlog_2(X * (X+5)) = 1\n2 ^ log_2(X * (X+5)) = 2^1\nX * (X+5) = 2\nX^2 + 5*X = 2\nX^2 + 5*X - 2 = 0\nX = (-5 +- sqrt(5^2 + 4*1*2)) / 2\nX = (-5 +- sqrt(33)) / 2\n\nSo, two possible answers are:\nX1 = (-5 + sqrt(33)) / 2\nX2 = (-5 - sqrt(33)) / 2\n\nAlso, logarithms have sense only if these inequality are true:\nX > 0\nX + 5 > 0\n\nX2 < 0, then X1 is the only possible answer.\n\nFinally:\nX = (sqrt(33) - 5) / 2
898977
log(2x+1)-(1/3)log(x+1)\n=log(2x+1) - log(  (x+1)^(1/3)  )\n=log( (2x+1) /   (x+1)^(1/3)  )
743256
First 4*log[x-2] = log[x-2]^4. \nThen -log[5] = log[1/5]\n\nThe sum of logs is the log of the product of the arguments and so this gives\n\n(1/2)*log[(x-2)^4 * (x+3) / 5]\n\nFinally the (1/2) becomes an exponent in the argument:\n\nlog[((x-2)^4 * (x+3) / 5)^(1/2)]
1060074
d (ln(x)) / dx = 1/x\n\nThe derivative of the natural log of x with respect to x equals 1 / x
833486
int1/x^2+int1/x^5\nx^-1/-1 +x^-4/-4+C\n-1/x-1/4x^4+C
496208
log(x) - log(1 + (x^2 + 1)^(1/2))\n\nWhere log is the natural logarithm, of course.
140923
The other property of logarithms is that Log_x A - Log_x B = Log_x (A/B)\n\n\nso you can combine this rule with the other one i told you about in your other question...\n\nthe only thing is that u are working in reverse order\n\n\nLn 2y/x= Ln 2+ Ln y - Ln x\n\n\nThese are properties, they should be tought to you as principals, without them you cant solve logs:D hope this helps ;) if you think it does pls gimme 10 pts:P
719380
Let y = x^(1/2)\n\nso y^2 = x,\n\ndx = 2y dy\n\nThe problem becomes:\n\nint [1 / (y + y^3)] 2y dy\n\nOr:\n\nint 2 [1 / (1 + y^2)] dy\n\nA recognizable integral. Integrate and substitute back x^1/2 = y
862518
When we need to ind the logarithms of number not of base ten, we can use log*6*x=logx/log6 so the answer is\n(log60)/(log6)-(log30)/(log6)
14407
This is an integration-by-parts problem. The answer is:\nx*ln(x)+C\n\nThis is how you calculate it:\n\nIntegral of ln(x)*dx\n\nu= ln(x)\ndu=1/x*dx\ndv=dx\nv=x\n\nSo by parts: uv-integral of vdu\n= ln(x)*x-integral of x*1/x*dx\n=x*ln(x)-integral of 1*dx\n=x*ln(x)-x+c
1106006
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.
347784
Taking natural logarithm to both sides\nLn(y)= Ln(x) * Ln(x) = (Ln(x))^2\nDifferentiating\n(1/y )dy/dx = (2/x) * (Ln(x))\ndy/dx = (2y/x) * (Ln(x))\nat the point (e,e),\ndy/dx = (2e /e) * Ln(e)\nso the slope = 2
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.
593752
log ( sqrt (x+ 1) *x^5)
1159338
Integrate using the method of substitution based on the fact that the derivative of log(y) + C = 1/y\n\nlet u = log(y) then du/dy = 1/y. So du = (1/y) dy\n\nUsing these substitutions we have:\n\nInt 1/y*log(y) dy  = Int 1/log(y)*(1/y) dy\n\n= Int 1/u du  = log(u)+C = log(logy)+C, since u = log(y)\n\nThis is correct since from the chain rule:\n\n d/dy log(logy)+C= 1/log(y)*d(logy)/dy= 1/(y*logy)
1000878
First let I(k,n)=int (log x)^k x^n dx where the integral goes from 0 to 1. Do an integration by parts with u=(log x)^k x^(n-1) and dv=x dx\nto get that I(k,n)=-k/(n+1). By induction, we then obtain that\nI(n,n)=(-1)^n n!/(n+1)^n I(0,n)\n=(-1)^n n!/(n+1)^(n+1).\n\nNow, write x^(ax)=exp(ax log x)=\nsum 1/n! (ax log x)^n=\nsum (1/n!) a^n x^n (log x)^n.\n\nNow integrate. Since the sum converges uniformly, the interchange of the sum and integral gives\nint x^(ax) dx=sum (1/n!) a^n int x^n (log x)^n dx\n= sum (1/n!) a^n I(n,n)\n=sum (-1)^n a^n /(n+1)^(n+1)\nas desired.