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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: stuart clark - Haziran 09, 2011, 05:17:36 öö
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(http://latex.codecogs.com/gif.latex?\mathbf{\displaystyle{\int\limits_1^e%20{\frac{{{e^x}\left(%20{x{{\ln%20}^2}x%20+%20{e^x}}%20\right)}}{{x{{\left(%20{{e^x}%20+%20\ln%20x}%20\right)}^2}}}dx}=%20}}$)
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Well, it took long to get the answer, I have tried several ways to do partial integration but couldnt see anything useful. My solution goes in a different way:
We have a function in the integral in the form of K(x) / M(x)2 where M(x) = (ex+lnx); Think of a solution in the form
f(x)/g(x), then its derivative would be (f'g-g'f)/g2 where M(x) = g(x). Assume it is in this form; if it is we can find it.
Take g(x) = ex+lnx, Then K(x) = exln2x + e2x/x which also equals f'g-g'f.
f'(x)(ex+lnx) - f(x)(ex+1/x) = exln2x + e2x/x
The above equation in bold, is the worst kind of differential equation :) Non-homogeneous with non-constant coefficients. What could f be? K(x) has exponential term with second degree, so, if f(x) does not have an exponential function inside, we cannot construct the left hand side with exp 2-deg. So, if there is an f(x), it must have the following format: f(x) = a(x)ex + b(x)
Let's substitute f(x):
(exa'(x) + exa(x) + b'(x))(ex+lnx) - (exa(x)+b(x))(ex+1/x) = K(x)
e2xa'(x) + ex lnx a'(x) + exa(x)lnx + exb'(x) + b'(x)lnx - exa(x)/x - bex - b(x)/x = K(x) = exln2x + e2x/x
So, if f(x) exists then e2xa'(x) = e2x/x because a(x) and b(x) are non-exponential.
Then a'(x) = 1/x and a(x) = lnx + C
Also -b(x)/x + b'(x)lnx = 0 (Non-exponential terms of both left and right hand side must be equal)
(-b(x)/x + b'(x)lnx)/ln2x = 0 --> (b(x)/lnx)' = 0 --> b(x) = C lnx where C is a constant
From ex terms(substitute a(x) and a'(x)) we have lnx/x + ln2x + b'(x) - lnx/x) = ln2x
--> b'(x) = 0 --> b(x) = C = C lnx --> C = 0, so b(x) must be 0
Then our f(x) function turns out to be f(x) = exlnx
So, the result of the indefinite integral is I(x) = f(x)/g(x) = exlnx/(ex+lnx)
Calculating at limits --> I(e) - I(1) = ee/(ee+1)
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A great job..
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Thanks a lot
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Thanks Moderator for nice solution.
Here is my solution.
(http://latex.codecogs.com/gif.latex?\displaystyle{I%20=%20\int\limits_1^e%20{\frac{{{e^x}\left(%20{x{{\ln%20}^2}x%20+%20{e^x}}%20\right)}}{{x{{\left(%20{{e^x}%20+%20\ln%20x}%20\right)}^2}}}dx}%20}=\int\limits_1^e\frac{e^x\left(\ln^2x+\frac{e^x}{x}%20\right%20)}{\left(e^x+\ln%20x%20\right%20)^2}dx$\\\\\\%20Now%20$\displaystyle\frac{e^x\left(\ln^2x+\frac{e^x}{x}%20\right%20)}{\left(e^x+\ln%20x%20\right%20)^2}=\frac{d}{dx}\left(\frac{f(x)}{e^x+\ln%20x}\right)=\frac{f^{'}(x)\left(e^x+\ln%20x\right)-f(x)\left(e^x+\frac{1}{x}\right)}{\left(e^x+\ln%20x\right)^2}$\\\\\\%20$\displaystyle%20f^{'}(x)-f^{'}(x)x\ln%20x=\frac{e^x}{x}-e^xx\ln^2x\\\\\\%20$\displaystyle%20f^{'}(x)=\frac{\frac{e^x}{x}-xe^x\ln^2x}{1-x\ln%20x}=\frac{e^x}{x}\cdot\frac{1-x^2\ln^2x}{1-x\ln%20x}$%20\\\\\\%20$So%20$f(x)%20=%20e^x\ln%20x+C$\\\\\\%20at%20$C%20=%200$%20,%20value%20of%20$f(x)%20=%20e^x\ln%20x$\\\\\\)
(http://latex.codecogs.com/gif.latex?{I%20=%20\int\limits_1^e%20{\frac{{{e^x}\left(%20{x{{\ln%20}^2}x%20+%20{e^x}}%20\right)}}{{x{{\left(%20{{e^x}%20+%20\ln%20x}%20\right)}^2}}}dx}%20}=\int\limits_1^e\frac{e^x\left(\ln^2x+\frac{e^x}{x}%20\right%20)}{\left(e^x+\ln%20x%20\right%20)^2}dx\\\\\\%20\displaystyle%20=%20\int\limits_1^e\frac{d}{dx}\left(\frac{f(x)}{e^x+\ln%20x}\right)dx=\left.\frac{e^x\ln%20x}{e^x+\ln%20x}\right|_1^e=\boxed{\frac{e^e}{e^e+1}})
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How did you connect second and third lines?