integrating (1/x)dx  by parts gives O = 1..... is it true?????
94569
Your integral needs either limits (from 1 to X, for example) or a constant added to it. With limits, the apparent contradiction vanishes. With an arbitrary constant, it can be accounted for.
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...
574891
im pretty sure that you will have to use the parts method to solve it. but the answer is\n\nx(log(1-ie^(ix))-log(1+ie^(ix))+i(Li(-ie^(ix)-Li(ie^(ix))\n\ni=imaginary unit\nlog=natural logarithm\nLi=polylogarithm\n\nyou can check my answer to see if its right because i also thought that it was wrong. i dont know how the imaginary units could have shown up but o well.
1092276
Let u = 3 - x^2, then du = -2x dx\nso (-1/2) du = x dx\nNow the integral is\n(-1/2)(1/u^2) du\n= (-1/2)(u^(-2)) du\n= (-1/2)u^(-1)/(-1) + C\n= (1/2)(u^(-1)) + C\n= 1/(2u) + C\n= 1/(2(3-x^2)) + C\nThere is no need to solve for x = sqrt(-u+3).  The x in the integral cancels with the x from du.
57930
The integration by parts formula is\nint(u*dv)=u*v-int(v*du)\n\nLet u=r and dv=e^(r/2)dr\n\nNow take the derivative of u with respect to r:\ndu=dr\n\nNow integrate dv\nint(dv)=int[e^(r/2)dr]\nv=[e^(r/2)]/(1/2)  --> The chain rule was used here. \n__=2e^(r/2)\nWhen you take the derivative you multiply by "what's inside" so when you integrate, you have to divide by it.\n\nNow plug into the formula\n2re^(r/2)-int(2e^(r/2)dr)\n\n=2re^(r/2)-2(e^(r/2))/(1/2)+C\n\n=2re^(r/2)-4(e^(r/2))+C\n\n=(r-2)*2e^(r/2)+C
850555
You can do it either way, but one way might be better depending on the limits of your integrals. \n\nIf you integrate with repect to x first, you do a substitution u=yx^2,\ndu=2xy dx to get an anti-derivative of (1/2)e^(x^2 y). The limits of integration on x will determine how hard the next integral (with respect to y) will be.\n\nIf you integrate first with respect to y, you have to do an integration by parts with u=xy and dv=e^(x^2 y) dy. Then du=xdy and v=(1/x^2) e^(x^2 y). You will still have to integrate v du to get an anti-derivative of\n(y/x)e^(x^2 y) - (1/x^3) e^(x^2 y).\nAgain, the limits of your integration will dictate how difficult the next integral will be.\n\nIf you have to change the order of integration, BE SURE YOU GET THE LIMITS OF INTEGRATION RIGHT.
952947
we recongsize that d/dx(x^3) = 3 x^2\n\nso let x^3-1 = t\n\n3x^2 dx = dt\nI am using int for integration sign below\n\n\nso int ( x^2/(x^3-1)^(1/2) dx\n= int( dt/3t^(1/2)\n\n= 1/3 int (dt/t(1/2)) = 1/3 int(t^-(1/2)\n\n1/3 (2 t^(1/2) + C using integral of t^n = t^(n+1)/(n+1)\n= 2/3 (x^3-1)^(1/2) + c
714362
Let u be 3e^x + 1, then du/dx = 3e^x .  So your integral amounts to \n\nintegral( (1/3 du/dx) u^(1/3)) = 1/3 integral(u^(1/3)du)\n\n= (1/3)*1/(1+(1/3))*u^(1 + 1/3) + C\n=(1/3)*(1/(4/3))*u^(4/3) + C\n= (1/3) * (3/4) * (3e^x+1)^(4/3) + C\n= (1/4)* (3e^x + 1)^(4/3) + C\n\nCheck by  differentiating:\n\nd/dx (1/4 (3e^x + 1)^(4/3) + C) = \n(4/3)* (1/4) * (3e^x + 1)^(1/3) * (d/dx (3e^x + 1)) + d/dx(C) =\n(4/12) * (3e^x + 1)^ (1/3) * 3e^x dx + 0 + 0 =\n(1/3) * (3e^x + 1)^(1/3) * 3e^x dx =\ne^x (3e^x + 1) dx\n\nwhich is what you started with.
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) ]
546126
intrgrate o e^(3x)=(1/3)e^(3x)
965897
y=1/x ?\n\nMultiply both sides by x:\nxy=1\n\nDivide both sides by y:\nx=1/y
798293
use integration by parts:\n\nu=x\ndv= cos(x) dx\n\ndu =dx\nv= sin(x)\n\nintegral of x cosx dx = xsinx - integral of sin(x) dx\n= x sinx + cos(x) + C\n[
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
735191
Writing this out in more detail:\n1/2 ∫cos (2x) dx\nu=2x, du=2 dx, dx=du/2\n1/2 ∫cos u/2 du\n1/4 ∫cos u du\n1/4 sin u + C\n1/4 sin (2x) + C\n\nThe u comes from the fact that you are integrating cos 2x, not cos x.\n\nOh, and in the second step, you should have ∫(1-cos (2x))/2 dx -- the way you wrote it literally means ∫1 - x (cos 2)/2 dx, which is absurd.
168841
Suppose that the limits on your integration are from 1 to X\n\nint{1/x dx from 1 to X} = (1-1)+int{1/x dx  from 1 to X}\n\nand one finds out that 0=0\n\nIf you want to perform an indefinite integral,\n\nint{udv}=uv-int(vdu} +C, where C is a constant.\n\nif u=1/x, and v=x, then choosing C=-1 makes this equality true.
1122081
It's called the integration by parts formula.\nInt u dv=uv- int v du\n\nBoth u and v need to be absolutely continuous for this to work. A continous derivative for both is plenty.
726082
Yes since ∫f(x) dx = ∫x dx\n= ½x² + c where c is ANY constant.