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The integral I am dealing with is below.

I need to find the closed-form expression of this integral.

$\int_0^\infty \ln(1+\frac{A}{1+B+Cx})\frac{e^{-x/M}}{M}dx$

Here, $A$, $B$, $C$ and $M$ are constants.

How can I do it in Mathematica?

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Mathematica can do it if you add assumption:

$Version
(*"12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)"*)

Integrate[Log[1 + a/(1 + b + c*x)]*Exp[-x/m]/m, {x, 0, Infinity},Assumptions -> {a > 0, b > 0, c > 0, m > 0}]

(*E^((1 + b)/(c m)) ExpIntegralEi[-((1 + b)/(c m))] - 
 E^((1 + a + b)/(c m)) ExpIntegralEi[-((1 + a + b)/(c m))] + 
 Log[(1 + a + b)/(1 + b)]*)

In LaTeX code:

$\int_0^{\infty } \frac{\ln \left(1+\frac{a}{1+b+c x}\right) \exp \left(-\frac{x}{m}\right)}{m} \, dx=e^{\frac{1+b}{c m}} \text{Ei}\left(-\frac{1+b}{c m}\right)-e^{\frac{1+a+b}{c m}} \text{Ei}\left(-\frac{1+a+b}{c m}\right)+\ln \left(\frac{1+a+b}{1+b}\right)$

for:$\{a>0,b>0,c>0,m>0\}$

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  • $\begingroup$ Here, E is Exp? $\endgroup$ – dipak narayanan Apr 26 at 15:59
  • $\begingroup$ @dipaknarayanan .Yes. $\endgroup$ – Mariusz Iwaniuk Apr 26 at 16:03

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