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Symbolic integral of an integrand containing derivative

How to calculate this integral in Mathematica?
integral[(-(df/dx)* x^n(x+2x^2))dx, 0, inf]
POSTED BY: Dianta Ginting
5 Replies
The above is not Mathematica syntax. It is also not Latex. It is free hand format and hard to read. It is not clear what you wrote. is "kt" a parameter? or is it "k*t". If it is "k*t", what is "k"? a constant? does it have value? but f is a function of "x" and not "t" so it does not make sense to have "k*t" in there. so may be then "kt" is the constant? but then does "kt" have a value? and what is "I=0"? Is "I" used somewhere else?  I do not see an "I" in the equation.  Is "^l/2" in your equation an "^1/2"?  Hard to read the "l" from a "1" on the screen. But you said that "I=0" before, so it can't be "^I" since that will make the whole thing 1 because anything to the power zero is 1 (except for 0^0), so it must be "1/2", but if so, why then say "I=0" before and not use it? You also say "n=0", then you have "x^n" in the equation. But "x^0" is 1. So why have it in there in first place?  You also have 1/1+Exp[x-1], but 1/1 is 1? so why have it there?

Why not use valid Mathematica code? If you do not know the syntax of Mathematica, you can try Wolfram|Alpha, it should accept natural human input similar to what you have above. Or you can post an image of the equation from the book so it is clear what it is.

This is my best guess:
m = 3/2;
n = 0;
a = 2;
kt = 1;
f[x_] := 1/1 + Exp[x - 1]/(kt)
Integrate[-f'[x]*x^n*(x + a*x^2)^m*((1 + 2*a*x)^2 + 2)^1/2, {x, 0,Infinity}]
But this does not converge.
POSTED BY: Nasser M. Abbasi
The function is
f(x) = 1/1+exp(x-1)/kt
 I should calculate the Fermi integral
F(x) = integral[(-(df/dx)* x^n*(x+a*x^2)^m*[(1+2*a*x)^2+2]^l/2dx, 0, inf]

I try to calculate in Mathematica, but the problem come with error.

Thank for you help.
POSTED BY: Dianta Ginting
Hello and welcome to the Wolfram Community! Please take a few minutes to read this tutorial about correct posting – especially of Mathematica code:

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POSTED BY: Moderation Team
If you mean you want to syntax, then
Integrate[-f'[x]*x^n (x + 2 x^2), {x, 0, Infinity}]
Then if you have f defined before that, say
f[x_] := Exp[-x]
Then Mathematica should reply with
ConditionalExpression[(5 + 2 n) Gamma[2 + n], Re[n] > -2]
POSTED BY: Nasser M. Abbasi
There is no way to know how to help you unless you provide the function f.
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