Can anyone give me a proof of how the integral of [1/(1=E^x)]dx = x-ln(1+E^x)?
408775
1/(1+e^x)=(1+e^x -e^x)/(1+e^x)=1-  e^x/(1+e^x)=\n1- (1+e^x)'/(1+e^x)\n& u'll integrate
920193
e^(xlnx )=x^x\nderivative is  1+ln x
882612
2x^2 ln(x) - x^2\n\nVerify by differentiating\n2x + 4x ln(x) - 2x = 4x ln(x)
835481
This depends strongly on how you define ln(x). If you define it as the inverse function of e^x and you know that the derivative of e^x is e^x, then use the formula for the derivative of an inverse function:\nd/dx[f^(-1) (x)]=1/f'[f^(-1) (x)].\nWith f(x)=e^x, f^(-1) (x)=ln(x) and we get\nd/dx[ln(x)]=1/e^(ln(x))=1/x.\n\nIf you define ln(x) as the integral from 1 to x of 1/t dt, then the fundamental theorem of calculus immediately gives the result.\n\nIf you only have the properties of logarithms and the definition of e as\nlim (1+ x)^(1/x)\nas x->0, \nthen the result is more complicated:\n\nLet f(x)=ln(x), then\nf(x+h)-f(x)=ln(x+h)-ln(x)\n=ln(1+h/x).\n\nThus, [f(x+h)-f(x)]/h =\n(1/h) ln(1+ h/x)=\n(1/x) (x/h) ln(1 + h/x)=\n(1/x) ln( [1+ h/x]^(x/h) ).\nThe term inside the ln is to be evaluated as h->0. But this gives, from the definition of e,\n(1/x) ln(e)=1/x.\n\nThe last method can be used to find the derivative of log_a (x)\nas \n(1/x) log_a (e).
1085682
Ln(e^x+1) = 0 by log laws this equals X+1 Ln e = 0 and Ln e cancels therefore X+1 = 0, x = -1\n\nIf the equation does not = 0 then you will be just left with ( x + 1)
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)
410987
You were on the right track with the u substitution:\n\nlet u = sqrt(1 + ln x)\nthen u^2 = 1 + ln x\nu^2 - 1 = ln x\n2u du = dx/x\n\nSubstitute in to the original integral:\nS 2u^2/(u^2 - 1) du = 2S u^2/(u^2 - 1) du\n\nI found in an old book that I have that\nS x^2/(x^2 - a^2) dx = x + (a/2)ln[(x - a)/(x + a)]\n(I know it works, but I can't seem to explain how my book came up with that one.  To prove it to yourself, take the derivative of the right side and it works out to the expression inside the integral on the left.)\n\nSo, \n= 2[u + (1/2)ln[(u – 1)/(u + 1)]\n\nAnd substituting for u,\n= 2sqrt(1 + ln x) + ln[(-1 + sqrt(1 + ln x))/(1+ sqrt(1 + ln x))]
179378
All you do is take the natural logarithm (ln) of both sides.\nBecause the ln of e to any power is just that power it becomes this:\nln (e^-x) = ln(6) \nThe ln sort of "cancels out" the e, leaving you with the power so...\n-x = ln(6)\nMultiple both sides by negative 1 and you're done, unless you need to approximate ln(6).
452294
Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
885845
dy/dx+2y=2, if y(0)=0\n\ndy/dx=2-2y=2(1-y)\ndy/(1-y)=2dx\nintegrate\n-ln(1-y)=2x+C{C is a constant}\nln(1-y)= -2x-C\ntake'e'\n1-y= e^(-2x-C)\ny= 1-e^(-2x-C)\ntake initial conditions\n0=1-e^(-C)\n>>>>>C=0\ntherefore,y=1-e^(-2x)\n\ncheck in original equation\ndy/dx=2e^(-2x)\n2y=2-2e^(-2x)\ndy/dx+2-2e^(-2x)=2\nsolution agrees with original \nequation,therefore it is \ncorrect
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|)
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.
1138866
I'll lead you in the most important part, but it is up to you to complete it.\n\nDerivative of x . y = x'.y + x.y'\n\nThen, you will have:\n\nf'(x) = (2x . ln x^2) + x^2 (ln x^2)'\n\nContinue from here, the only problem that you may have is finding the derivative of ln y , which you can see in the link below.
773418
derivative of e^u = e^u du\nderivative of ln(u) = du/u\na^u=e^{uln(a)}, so the derivative of a^u =  e^{uln(a)} ln(a) = ln(a) a^u\ns
275631
One way to see this is to write x^(1-x) as exp((1-x)ln x) and use the chain rule. Only one person has the correct answer, art_colls. The power rule *does not apply* since the exponent is a function also!
84808
multiply and divide by (1 - sinx)\n\nyou get (1 - sin^2(x))/(cosx(1-sinx)) that is cos^2(x)/(cosx(1-sinx)) that is cosx/(1-sinx)\n\nnow at the denominator you have a function and at the numerator you have the derivative of that function, save for the sign. \nyour integral is \n-ln(|1-sinx|) + c\n\n---- \n\nproof that we're all saying the same thing:\n- ln |1-sinx| = ln (|1 - sinx|^-1) =\n\n= ln |1/(1-sinx)| ---> divide and multiply again for \n(1 + sinx)  ----> = ln |(1 + sinx)/cos^2(x)|\n\n= ln |(secx + tanx)/cosx| = ln|(secx + tanx)secx| = \n= ln |secx + tanx| + ln|secx|\n\nSo you have many ways to write your answer: you can choose the one which you prefer\n- by elegance\n- by minimization of spread error in calculating it\n- by whatever you like\n\nhave a nice day
755844
1 correct \n\n2 ∫x/e^x dx \n= ∫x*e(-x) dx\n= -xe^(-x) - ∫-e^(-x). 1.dx\n= -xe^(-x) - e^(-x) + c\n\nSo∫[x = 1 to ∞] x/e^x dx = lim t →∞ (-te^(-t) - e^(-t)) - (-1/e - 1/e)\n= 2/e (NB lim t →∞ (-te^(-t)) = o by d'Hopitals rule)\n\n3 correct\n\n4. Use (n² + 1)/[e^n * (n + 1)²] \n=  [(n² + 1)² - 2n]/[e^n * (n + 1)²]\n= 1/e^n - 2n/e^n(n + 1)² \nNow 1/e^n < 1/3^n and  2n/e^n(n + 1)² < 2n/3^n*n² \n= 2/n*3^n \nSo both parts converge so the difference converges\n\nNOTE: It is many, many years since I have done these BUT I am confident that I am correct ... Maybe you can tell me if that is so!