integral

[stuart clark]

[senior]

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) = (e^x+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) = e^x+lnx, Then K(x) = e^xln²x + e^2x/x which also equals f'g-g'f.
f'(x)(e^x+lnx) - f(x)(e^x+1/x) = e^xln²x + e^2x/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)e^x + b(x)
Let's substitute f(x):
(e^xa'(x) + e^xa(x) + b'(x))(e^x+lnx) - (e^xa(x)+b(x))(e^x+1/x) = K(x)
e^2xa'(x) + e^x lnx a'(x) + e^xa(x)lnx + e^xb'(x) + b'(x)lnx - e^xa(x)/x - be^x - b(x)/x = K(x) = e^xln²x + e^2x/x
So, if f(x) exists then e^2xa'(x) = e^2x/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 e^x 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) = e^xlnx
So, the result of the indefinite integral is I(x) = f(x)/g(x) =  e^xlnx/(e^x+lnx)
Calculating at limits --> I(e) - I(1) = e^e/(e^e+1)

[proble_m]

A great job..

[senior]

Thanks a lot

[stuart clark]

Thanks Moderator for nice solution.

Here is my solution.

[senior]

How did you connect second and third lines?