Integral of x^2 *ln (x+1)?
538185
Let u = ln(x+1) and dv = x^2dx, then\ndu = (1/(x+1))dx and v = (1/3)x^3\nuv- int(vdu)\n= (1/3)x^3ln(x+1) - (1/3)int(x^3/(x+1)dx)\nDivide x^3/(x+1) using long or synthetic division to get\nx^2 - x + 1 - 1/(x+1)\nint(x^2 - x + 1 - 1/(x+1))\n= (1/3)x^3 - (1/2)x^2 + x - ln(x+1)\n= (1/3)x^3ln(x+1) - (1/3)((1/3)x^3 - (1/2)x^2 + x - ln(x+1))\n= (1/3)x^3ln(x+1) - (1/9)x^3 + (1/6)x^2 - (1/3)x + \n(1/3)ln(x+1)
599364
x^2 + x = x(x+1)\n\nIt can be simplified to x(x+1).
566425
Summation from x=1 to x=n of x^2
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
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)
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)
242020
i'll do it by 2 ways \nthe first one if u know what integration by parts is:\n(int=integral)\nint[udv]=uv-int[vdu]\nlet u=lnx =>du=dx/x\n& let dv=xdx => v=(x^2)/2\nso we'll have\nint[xlnx]=((x^2)/2) lnx -1/2int[xdx]\n=((x^2)/2)lnx -((x^2)/4)+ Constante\nthe second method is\ntry derivating (x^2)lnx u'll get [(x^2)lnx]'=2xlnx +x\nthen integrate int{[(x^2)lnx]dx}=(x^2)lnx\n&int[x]=(x^2)/2\nso u'll get (x^2)lnx=2int[xlnxdx]+(x^2)/2\nso int[xlnxdx]=((x^2)/2)lnx-(x^2)/4+ C
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
903727
-1 / ( 20 ( ln( 10 + e ) * ( 10 + e )^(10 * x^2 ) ) )
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) ]
883869
http://integrals.wolfram.com/index.jsp\nhas a beaut integrator\n\nIntegrate[Log[Cos[x]], x] == \nI  2              (2 I) x\n- x  - x Log[1 + E       ] + \n2\n \n  x Log[Cos[x]] + \n \n  I              (2 I) x\n  - PolyLog[2, -E       ]\n  2\n\nI tried to paste the devilish-looking solution
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
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
900298
We know that F(x) = ∫ln³ x/x dx + C. So we need to find that integral, solve for the constant of integration, and then plug 9 into the resulting formula. First, the integral:\n\n∫ln³ x/x dx\nLet u=ln³ x, v=ln x, du=3 ln² x/x dx, dv = 1/x dx. Integrating by parts:\n∫ln³/x dx = ln^4 x - ∫3 ln³ x/x dx + C\nAdding 3∫ln³ x/x dx to both sides:\n4∫ln³/x dx = ln^4 x + C\nDividing by 4:\n∫ln³/x dx = ln^4 x + C\n\nNow, solving for C:\n\nF(1) = ln^4 1 + C\nF(1) = 0^4 + C\nF(1) = C\nBut F(1) = 0, so C=0.\n\nNow, solving for F(9):\n\nF(9) = ln^4 9 = (ln 9)^4 ≈ 23.30761
1051703
collect like terms on one side\nxdx/(1+x^2)=dy/[(1+y^2)*y]\n\n1/2* 2x/(1+x^2)= dy/y-1/2*2y/(1+y^2)\n\nBy integrating each side,\n1/2*ln(1+x^2)\n=ln|y|-1/2*ln(1+y^2)\n\nuse log rules, take square of both sides,\n\nC*y^2 / (1+y^2)=1+x^2\n\ny=sqrt([1+x^2]/[C-1-x^2])
68076
god man.  what kind of horrible painful calculus class are you taking?\n\nfirst integrate over dx.  the integral of x^y dx is just a power integral, so you have y*x^(y-1).  the integral of y^x dx is just an exponential, so you have y^x/ln y.\n\nnow integrate y*x^(y-1) + y^x/ln y over dy.  and there, my friend, is where i stop because frankly i can't be arsed to do something so brain-curdling.  actually the first term doesn't look so bad, i think if you integrate it by parts it should come out in one step.  don't really want to think about doing the 2nd one.\n\nyou might consider trying some kind of fancy thing like transforming to polar coordinates or something.  might make it easier.\n\nbtw you should know about this site:\nhttp://integrals.wolfram.com/