Explain the answer to this integral:  The integral of 8/(1+5t) with respect to t is 8/5 ln (1+5t)?
403275
Look, to count this integral, you must make a subsitution:\n\nu = 1+5t\ndu = 5 dt\ndt = du/5\n\nint 8/(1+5t) dt =\n= int 8/u du/5 =\n= 8/5 * int (1/u) du =\n= 8/5 * ln u + C =\n= 8/5 * ln(1+5t) + C
333624
(ln x)^7=(7*(lnx)^6)*(1/x)=7*(lnx)^6/x
763343
Heartsense is right.  I'm chiming in to say if you mean the product of 5/6 * 6/7 * 7/8 * 8/9 * ..., then that product tends to 0 since the denominator of the nth term cancels out with the numerator of the (n+1)st term, so as n approaches infinity, you get 5 over infinity (not really "over infinity" but you know what I mean) = 0.
1090386
What?\nIf u=x^x\ndu/u=(1+ln(x))dx\n\nInteral  exp(x^x)dx=integral exp(u)du\n\nAnalytical solution for this indefinite integral is dificult.
641969
log 35 = log (7 * 5) = log 7 + log 5\n\nlog 8 = log (2^3) = 3 log 2 = 3 * 0.3010\n\nlog 5 is already given isn't it. \n\nand by the way you have written the value of log 7 wrongly. the value that you have given is of log 3.
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
719088
Substitute u=pi/2-x. Then tan(x)=cot(u), du=-dx and the limits of integration are reversed. So, if we let I_n be the value of the integral, we also have\nI_n is the integral of 1/(1+cot^n (u))du=\nintegral of tan^n(u)/(1+tan^n (u)) du.\n\nNow replace u by x and add to the original to get that\n2I_n =int (1+tan^n (x))/(1+tan^n(x)) dx=pi/2.\n\nThus I_n=pi/4.
883154
I assume that's suppose to be an integral symbol.\n\nf(x) = [1/2 t²] from 2 to x, or 1/2 x² - 1\n\nThen f'(x) = x, and f''(x) = 1.\n\nDidn't really need to do the integration...
1102691
f(t) = 5 * t^4 * exp(-7*t) is of the form:\n\nf(x) = x^n * exp(-a*x)\n\nFunctions of this form, with n = an integer > 0 have Laplace transforms equal to:\n\nL[f(x)] (s) = n!/((s + a)^(n+1))\n\nFor your function, n = 4 and a = 7, so:\n\nL[f(t)] (s) = 5 * 4!/(s+7)^5 = 120/(s+7)^5\n\nSee source for a table of transforms.
876847
84885444445774587459/8 = 10610680555721823432.375\nThat + 156145355545 = 10610680711867178977.375\nThat - 5 = 10610680711867178972.375\n\nAnswer is 10610680711867178972.375
114706
Change of variable t=Tan[x]\nwe have dx=(1+Tan[x]^2)dt\n\nSubstitute:\nIntegral[(1+Tan[x]^2)\n/(1-Tan[x]^2)]dx\n=\nIntegral[dt/(1-t^2)]\n=...\n\nusing partial fraction decomposiotion\n2/(1-t^2)=1/(t+1)-1/(t-1)\n\n...=\n(1/2)(Integral[dt/(t+1)]\n-Integral[dt/(t-1)])\n=\n(1/2)(Ln[Abs[t+1]]\n-Ln[Abs[t-1]])\n=\n(1/2)Ln[Abs[(t+1)/(t-1)]]\n\nreplacing back t we have\n\n(1/2)Ln[Abs[(Tan[t]+1)\n/(Tan[t]-1)]]\nplus C\n\nNote that Abs is needed!\nIf it is used then we can write the result as you requested:\n\n(1/2)Ln[Abs[(1+Tan[t])\n/(1-Tan[t])]]\nplus C
791165
ln (x-1) + ln (x-5) = ln(3x+7)\nln( (x-1)(x-5) ) = ln(3x+7)\nso\n(x-1)(x-5)  = (3x+7)\nx^2 - 6x + 5 = 3x + 7\nx^2 -9x -2 =0\nx= (9+- sqrt( 81 -4(-2)) )/2\nx = ( 9 +- sqrt(89) )/2\n`
32210
integral of sec(x) = ln(|sec(x) + tan(x)|)\n\n\nsec(x) = (sec(x)(tan(x) + sec(x))/(tan(x) + sec(x))\n\n= (sec(x)tan(x) + sec(x)^2)/(tan(x) + sec(x))\n\nif u = (tan(x) + sec(x)), then du = (sec(x)tan(x) + sec(x)^2) \n\nso the integral is the integral of du/u = ln(|u|)
173783
I believe that the integral of 1/x is equal to the natural log of x + C.  (ln x + C).  Usually you take the antiderivative of the function but there are exceptions to the rule, like 1/x.
337080
when you integrate 1/x you get In |x|\n\nso the answer is -1/2 * In |x| + constant term\n\n\nYou can use substitution, but there's not much point, because you'll get 1/u then (which is the same thing). 1/x is a well known integrand. You'd be better just memorising that the integral of 1/x = In(x)
963851
The question doesn't make sense: what is T(n/4) supposed to mean when n = 2 or 3 or 5 ...??
432536
Oops………..!\n\ndac2chari\n\n\n u = 1/(ln t) Then du = t dt ………………..?  How that come …!\n\nIf u = 1/(ln t) then  u = [ ln t ] ^-1 \n\nAnd  du = - [ln t ] ^-2 [ 1/t ] dt = dt / [-t (ln t ) ^2] … please be careful.\n------------------------------------------------------------------------------------\n\nOK… this integral can’t carry out using analytical method; however, the flowing one can be solve\n\n∫ dt / [t ln(t)] \n\nu = ln (t)\ndu =  dt / t  \n\nGives\n\n∫ dt / [t ln(t)]  = ∫ du/ u  ln(u)\n\nSubstitute for u\n\n∫ dt / [t ln(t)]  = ln [ ln (t) ]